Multiple Choice Questions on Limits and Continuity Pdf

Latest Limit and Continuity MCQ Objective Questions

Limit and Continuity MCQ Question 1:

What is \(\lim\limits_{x \to 0}\frac{sinx~+~log(1-x)}{x^2}\)  equal to?

1. 0
2. 1
3. -1/2
4. 1/2

Answer (Detailed Solution Below)

Option 3 : -1/2

Concept:

L-Hospital Rule: Let f(x) and g(x) be two functions

Suppose that we have one of the following cases,

1. \(\lim\limits_{x \to a} \frac{f(x)}{g(x)} =\frac{0}{0}\)
2. \(\lim\limits_{x \to a} \frac{f(x)}{g(x)} =\frac{\infty }{\infty }\)

Then we can apply L-Hospital Rule ⇔\(\lim\limits_{x \to a} \frac{f(x)}{g(x)} =\lim\limits_{x \to a} \frac{f'(x)}{g'(x)}\)

Note: We have to differentiate both the numerator and denominator with respect to x unless and until

\(\lim\limits_{x \to a} \frac{f(x)}{g(x)} =l\neq \frac{0}{0}\) where l is a finite value.

Calculation:

\(\lim\limits_{x \to 0}\frac{sinx~+~log(1-x)}{x^2}\)

Form of limit\(\frac{0}{0}\)

Applying the L-Hospital rule,

\(\Rightarrow \displaystyle \lim_{ x\to 0}\frac{cosx+\frac{1}{1-x}(-1)}{2x}\)

\(\Rightarrow\displaystyle \lim_{ x\to 0}\frac{cosx-\frac{1}{1-x}}{2x}\)

\(\Rightarrow\displaystyle \lim_{ x\to 0}\frac{cosx-({1-x)^{-1}}}{2x}\)

Now Again Form of limit\(\frac{0}{0}\)

Applying the L-Hospital rule,

\(\Rightarrow\displaystyle \lim_{ x\to 0}\frac{-sinx-\frac{(1-x)-(-1)}{(1-x)^{2}}}{2}\)

\(\Rightarrow\displaystyle \lim_{ x\to 0}\frac{-sin~x-[(-1)(1-x)^{-2}(-1)]}{2}\)

\(\Rightarrow\displaystyle \lim_{ x\to 0}\frac{-sin~x-[(1-x)^{-2}]}{2}\)

\(\Rightarrow\displaystyle \lim_{ x\to 0}\frac{-sin~0-[(1-0)^{-2}]}{2}\)

=\(\frac{0-1}{2}\)

= -1/2

Limit and Continuity MCQ Question 2:

The value of k which makes \(f\left( x \right)\; = \;\left\{ {\begin{array}{*{20}{c}} {\frac{sin~5x}{x}\;,x \ne 0}\\ {k\;,x\; = \;0} \end{array}} \right.\)  continuous at x = 0, is

1. 2
2. 3
3. 4
4. 5

Answer (Detailed Solution Below)

Option 4 : 5

Concept:

For Continuity of the function f(x), Left hand limit = Right hand limit

i.e.\(\displaystyle \lim_{ x\to 0^{-}}\)f(x) =\(\displaystyle \lim_{ x\to 0^{+}}\)f(x) =\(\displaystyle \lim_{ x\to 0^{}}f(x)\)

\(\displaystyle \lim_{ x\to 0^{}}\frac{sinx}{x}=1\).........(Equation 1)

Application:

For Continuity of the function f(x), Left hand limit = Right hand limit

i.e.\(\displaystyle \lim_{ x\to 0^{-}}\)f(x) =\(\displaystyle \lim_{ x\to 0^{+}}\)f(x) =\(\displaystyle \lim_{ x\to 0^{}}f(x)\)

\(\displaystyle \lim_{ h\to 0^{}}\frac{sin5 (0-h)}{0-h}=\displaystyle \lim_{ h\to 0^{}}\frac{sin5 (0+h)}{0-+h}\)=\(\displaystyle \lim_{ x\to 0^{}} f(x)\)

\(\displaystyle \lim_{ h\to 0^{}}\frac{sin5 (0-h)}{0-h} \times \frac{5}{5}=\displaystyle \lim_{ h\to 0^{}}\frac{sin5 (0+h)}{0+h}\times\frac{5}{5}\) = k

\(5\displaystyle \lim_{ h\to 0^{}}\frac{sin5h}{5h}\) =\(5\displaystyle \lim_{ h\to 0^{}}\frac{sin5h}{5h}\) = k (Using equation 1)

5(1) = k

\(\Rightarrow\)k=5

Limit and Continuity MCQ Question 3:

What is \(\lim\limits_{x \to 1}\frac{2(1-cosx)}{x^2}\)  equal to?

1. \(4sin^2\frac{1}{3}\)
2. \(4sin^2\frac{1}{2}\)
3. \(sin^2\frac{1}{2}\)
4. \(0\)

Answer (Detailed Solution Below)

Option 2 : \(4sin^2\frac{1}{2}\)

Concept:

1-cosx=2\(sin^2\frac{x}{2}\)

Calculation:

In the given question, We have\(\displaystyle \lim_{ x \to 1} \frac{2(1-cosx)}{x^2}\)i.e.

\(\Rightarrow\) \(\displaystyle \lim_{ x \to 1} \frac{2(1-cosx)}{x^2}\)=\(\displaystyle \lim_{ x\to 1} \frac{(2)2sin^2\frac{x}{2}}{x^2}\)

\(\Rightarrow\) \(\displaystyle \lim_{ x \to 1} \frac{2(1-cosx)}{x^2} =\) \(\displaystyle \lim_{x \to 1} \frac {sin^2\frac{x}{2}}{\frac{x^2}{2^2}}\)

\(\Rightarrow\) \(\displaystyle \lim_{ x \to 1} \frac{2(1-cosx)}{x^2} =\) \(\frac{sin^2\frac{1}{2}}{\frac{1^2}{2^2}}\)

\(\Rightarrow\) \(\displaystyle \lim_{ x \to 1} \frac{2(1-cosx)}{x^2} =\) \(4sin^2\frac{1}{2}\)

Limit and Continuity MCQ Question 4:

The point(s), at which the function f given by f(x) = \(\left\{ \begin{array}{l} \frac{x}{{|x|}},x < 0\\ - 1,x \ge 0 \end{array} \right.\)  is continuous, is/are :

1. x ∈ R
2. x = 0
3. x ∈ R - {0}
4. x = -1 and 1

Answer (Detailed Solution Below)

Option 1 : x ∈ R

Concept:

A function f(x) is continuous at x = a, if\(\lim_{x\to a^{-}}\) f(x) =\(\lim_{x\to a^{+}}\)f(x) = f(a).

Explanation:

Given, f(x) =\(\left\{ \begin{array}{l} \frac{x}{{|x|}},x < 0\\ - 1,x \ge 0 \end{array} \right.\)

 We know that |x| = x when x > 0

and |x | = - x when x < 0

f(x) =\(\left\{ \begin{array}{l} \frac{x}{{(-x)}},x < 0\\ - 1,x \ge 0 \end{array} \right.\)

⇒ f(x) =\(\left\{ \begin{array}{l} -1,x < 0\\ - 1,x \ge 0 \end{array} \right.\)

Therefore, f(x) = -1

The function f(x) is a constant function, and a constant function is continuous for all x ∈ R.

The correct answer is option (1).

Limit and Continuity MCQ Question 5:

What is \(\displaystyle\lim_{h \rightarrow 0} \frac{\sin^2(x+h)−\sin^2x}{h}\)  equal to ?

1. sin 2  x
2. cos 2  x
3. sin 2x
4. cos 2x

Answer (Detailed Solution Below)

Option 3 : sin 2x

Concept:

L-Hospital Rule: Let f(x) and g(x) be two functions

Suppose that we have one of the following cases,

I.\(\lim\limits_{x \to a} \frac{f(x)}{g(x)} =\frac{0}{0}\)

II.\(\lim\limits_{x \to a} \frac{f(x)}{g(x)} =\frac{\infty }{\infty }\)

Then we can apply L-Hospital Rule ⇔\(\lim\limits_{x \to a} \frac{f(x)}{g(x)} =\lim\limits_{x \to a} \frac{f'(x)}{g'(x)}\)

Note: We have to differentiate both the numerator and denominator with respect to x unless and until

\(\lim\limits_{x \to a} \frac{f(x)}{g(x)} =l\neq \frac{0}{0}\) where l is a finite value.

Solution:

\(\displaystyle\lim_{h \rightarrow 0} \frac{\sin^2(x+h)−\sin^2x}{h}\)

Form of limit\(\frac{0}{0}\)

Applying the L-Hospital rule,

=\(\displaystyle \lim_{h \to 0}\frac{2sin(x+h)cos(x+h)-0}{1}\)

= 2sin(x + 0)cos(x + 0)

= 2sinx.cosx

= sin2x

∴ The correct option is (3)

Top Limit and Continuity MCQ Objective Questions

Find the value of\(\rm \displaystyle \lim_{x \rightarrow \infty} 2x \sin \left(\frac{4} {x}\right)\)

1. 2
2. 4
3. 8
4. \(\frac 1 2\)

Answer (Detailed Solution Below)

Option 3 : 8

Concept:

\(\rm \displaystyle \lim_{x \rightarrow 0} \frac{\sin x}{x} = 1\)

Calculation:

\(\rm \displaystyle \lim_{x \rightarrow \infty} 2x \sin \left(\frac{4} {x}\right)\)

=\(\rm 2 \times \displaystyle \lim_{x → ∞} \frac{\sin \left(\frac{4} {x}\right)}{\left(\frac{1}{x} \right )}\)

=\(\rm 2 \times \displaystyle \lim_{x → ∞} \frac{\sin \left(\frac{4} {x}\right)}{\left(\frac{4}{x} \right )} \times 4\)

Let\(\rm \frac {4}{x} = t\)

If x → ∞ then t → 0

=\(\rm 8 \times\displaystyle \lim_{t \rightarrow 0} \frac{\sin t}{t} \)

= 8 × 1

= 8

What is the value of\(\mathop {{\rm{lim}}}\limits_{x \to 0} \;\;\frac{{{{\left( {1 - \cos2x} \right)}^2}\;}}{{{x^4}}}\)

1. 1
2. 8
3. 4
4. 0

Answer (Detailed Solution Below)

Option 3 : 4

Concept:

• 1 - cos 2θ = 2 sin2 θ
• \(\mathop {{\rm{lim}}}\limits_{x \to 0} \;\;\frac{{\sin x}}{x} = 1\)

Calculation:

\(\mathop {{\rm{lim}}}\limits_{x \to 0} \;\;\frac{{{{\left( {1 - \cos2x} \right)}^2}\;}}{{{x^4}}}\)

=\(\;\mathop {{\rm{lim}}}\limits_{x \to 0} \;\;\frac{{{{\left( {2{{\sin }^2}x} \right)}^2}}}{{{x^4}}}\)          (1 - cos 2θ = 2 sin² θ)

=\(\;\mathop {{\rm{lim}}}\limits_{x \to 0} \;\;\frac{{4{{\sin }^4}x}}{{{x^4}}}\)

=\(\mathop {\lim }\limits_{x \to 0} 4\; × \;{\left( {\frac{{\sin x}}{x}} \right)^4}\)

= 4 × 1 = 4

Evaluate\(\rm \mathop {\lim }\limits_{x\rightarrow \infty} \frac{x}{\sqrt{1+2x^2}}\)

1. 0
2. 1
3. \(\frac{1}{\sqrt 2}\)
4. \(\frac 1 2\)

Answer (Detailed Solution Below)

Option 3 : \(\frac{1}{\sqrt 2}\)

Calculation:

We have to find the value of\(\rm \mathop {\lim }\limits_{x\rightarrow ∞} \frac{x}{\sqrt{1+2x^2}}\)

\(\rm \mathop {\lim }\limits_{x\rightarrow ∞} \frac{x}{\sqrt{1+2x^2}}\)       [Form\(\frac{∞}{∞}\)]

This limit is of the form\(\frac{∞}{∞}\), Here, We can cancel a factor going to ∞  out of the numerator and denominator.

\(\rm \mathop {\lim }\limits_{x\rightarrow ∞} \frac{x}{\sqrt{1+2x^2}}\)

=\(\rm \mathop {\lim }\limits_{x\rightarrow \infty} \frac{x}{x\sqrt{\frac{1}{x^2}+2}}\)

Factor x becomes ∞ at x tends to ∞, So we need to cancel this factor from numerator and denominator.

=\(\rm \mathop {\lim }\limits_{x\rightarrow \infty} \frac{1}{\sqrt{\frac{1}{x^2}+2}}\)

=\(\frac{1}{\sqrt{\frac{1}{\infty^2}+2}}=\frac{1}{\sqrt{0+2}}=\frac{1}{\sqrt 2}\)

Evaluate\(\rm \mathop {\lim }\limits_{x\rightarrow 0} \frac{\log (1+2x)}{\tan 2x}\)

1. -1
2. 1
3. 2
4. 4

Answer (Detailed Solution Below)

Option 2 : 1

Concept:

\(\rm \mathop {\lim }\limits_{x\; \to \;a} \left[ {\frac{{f\left( x \right)}}{{g\left( x \right)}}} \right] = \frac{{\mathop {\lim }\limits_{x\; \to \;a} f\left( x \right)}}{{\mathop {\lim }\limits_{x\; \to \;a} g\left( x \right)}},\;provided\;\mathop {\lim }\limits_{x\; \to a} g\left( x \right) \ne 0\)

\(\rm \mathop {\lim }\limits_{x\; \to 0} {\frac{{\tan x}}{x}} = 1\)

\(\rm \mathop {\lim }\limits_{x\; \to 0} {\frac{{\log (1+x)}}{x}} = 1\)

Calculation:

\(\rm \mathop {\lim }\limits_{x\rightarrow 0} \frac{\log (1+2x)}{\tan 2x}\)

\(\rm = \mathop {\lim }\limits_{x\rightarrow 0} \frac{\frac{\log (1+2x)}{2x} \times 2x}{\frac{\tan 2x}{2x} \times 2x}\\= \frac{\mathop {\lim }\limits_{x\rightarrow 0}\frac{\log (1+2x)}{2x} }{\mathop {\lim }\limits_{x\rightarrow 0}\frac{\tan 2x}{2x} }\)

As we know\(\rm \mathop {\lim }\limits_{x\; \to 0} {\frac{{\tan x}}{x}} = 1\) and\(\rm \mathop {\lim }\limits_{x\; \to 0} {\frac{{\log (1+x)}}{x}} = 1\)

Therefore,\(\rm \mathop {\lim }\limits_{x\; \to 0} {\frac{{\tan 2x}}{2x}} = 1\) and\(\rm \mathop {\lim }\limits_{x\; \to 0} {\frac{{\log (1+2x)}}{2x}} = 1\)

Hence\(\rm \mathop {\lim }\limits_{x\rightarrow 0} \frac{\log (1+2x)}{\tan 2x} = \frac 1 1=1\)

Evaluate\(\rm \mathop {\lim }\limits_{x\rightarrow \infty} \frac{x^2}{{1+x^2}}\)

1. 0
2. 1
3. 2
4. \(\frac 12 \)

Answer (Detailed Solution Below)

Option 2 : 1

Calculation:

We have to find the value of\(\rm \mathop {\lim }\limits_{x\rightarrow \infty} \frac{x^2}{{1+x^2}}\)

\(\rm \mathop {\lim }\limits_{x\rightarrow \infty} \frac{x^2}{{1+x^2}}\)       [Form\(\frac{∞}{∞}\)]

This limit is of the form\(\frac{∞}{∞}\), Here, We can cancel a factor going to ∞  out of the numerator and denominator.

\(\rm \mathop {\lim }\limits_{x\rightarrow \infty} \frac{x^2}{{1+x^2}}\)

=\(\rm \mathop {\lim }\limits_{x\rightarrow \infty} \frac{x^2}{x^2\left({\frac {1}{x^2}+1}\right)}\)

Factor x² becomes ∞ at x tends to ∞, So we need to cancel this factor from numerator and denominator.

=\(\rm \mathop {\lim }\limits_{x\rightarrow \infty} \frac{1}{\left({\frac {1}{x^2}+1}\right)}\)

=\(\frac{1}{{\frac{1}{\infty^2}+1}}=\frac{1}{{0+1}}=1\)

Examine the continuity of a function f(x) = (x - 2) (x - 3)

1. Discontinuous at x = 2
2. Discontinuous at x = 2, 3
3. Continuous everywhere
4. Discontinuous at x = 3

Answer (Detailed Solution Below)

Option 3 : Continuous everywhere

Concept:

• We say f(x) is continuous at x = c if

LHL = RHL = value of f(c)

i.e.,\(\mathop {\lim }\limits_{{\rm{x}} \to {{\rm{c}}^ - }} {\rm{f}}\left( {\rm{x}} \right) = \mathop {\lim }\limits_{{\rm{x}} \to {{\rm{c}}^ + }} {\rm{f}}\left( {\rm{x}} \right) = {\rm{f}}\left( {\rm{c}} \right)\)

Calculation:

\(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} {\rm{f}}\left( {\rm{x}} \right)\; = \;\mathop {{\rm{lim}}}\limits_{{\rm{x}} \to {\rm{a}}} \left( {{\rm{x}} - 2} \right)\left( {{\rm{x}} - 3} \right)\)            (a ϵ Real numbers)

\(= \mathop {{\rm{lim}}}\limits_{{\rm{x}} \to {\rm{a}}} {\rm{\;}}{{\rm{x}}^2} - 3{\rm{x}} - 2{\rm{x}} + 6\)

\(= \mathop {{\rm{lim}}}\limits_{{\rm{x}} \to {\rm{a}}} {\rm{\;}}{{\rm{x}}^2} - 5{\rm{x}} + 6\)

\(= {{\rm{a}}^2} - 5{\rm{a}} + 6\)

∴ f(x) = f(a), So continuous at everywhere

Important tip:

Quadratic and polynomial functions are continuous at each point in their domain

\(\mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{\sin \left( {{{\rm{x}}^3}} \right)}}{{{{\rm{x}}^2}{\rm{\;sinx}}}}\) is equal to

1. ∞
2. 0
3. 1
4. Can't say

Answer (Detailed Solution Below)

Option 3 : 1

Concept:

• \(\mathop {\lim }\limits_{{\rm{\alpha }} \to 0} \frac{{\sin \left( {\rm{\alpha }} \right)}}{{\rm{\alpha }}} = 1\) and \(\mathop {\lim }\limits_{{\rm{\alpha }} \to 0} \frac{{\rm{\alpha }}}{{\sin \left( {\rm{\alpha }} \right)}} = 1\)
• \(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \left[ {{\rm{f}}\left( {\rm{x}} \right){\rm{\;}} \pm {\rm{\;g}}\left( {\rm{x}} \right)} \right] = {\rm{\;}}\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} {\rm{f}}\left( {\rm{x}} \right){\rm{\;}} \pm \mathop {{\rm{lim}}}\limits_{{\rm{x}} \to {\rm{a}}} {\rm{\;g}}\left( {\rm{x}} \right)\)
• \(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \left[ {\frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}}} \right]{\rm{\;}} = \frac{{\mathop {{\rm{lim}}}\limits_{{\rm{x}} \to {\rm{a}}} {\rm{\;f}}\left( {\rm{x}} \right)}}{{\mathop {{\rm{lim}}}\limits_{{\rm{x}} \to {\rm{a}}} {\rm{\;g}}\left( {\rm{x}} \right)}}\)

Calculation:

To find: \(\mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{\sin \left( {{{\rm{x}}^3}} \right)}}{{{{\rm{x}}^2}{\rm{\;sinx}}}} = ?\)

Multiply and divide by x,

\(\mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{{\rm{x}}\sin \left( {{{\rm{x}}^3}} \right)}}{{{{\rm{x}}^3}{\rm{\;sinx}}}} = \mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{\sin \left( {{{\rm{x}}^3}} \right)}}{{{{\rm{x}}^3}}}\frac{{\rm{x}}}{{{\rm{sinx}}}}\)

\(= \mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{\sin \left( {{{\rm{x}}^3}} \right)}}{{{{\rm{x}}^3}}} \cdot \mathop {{\rm{lim}}}\limits_{{\rm{x}} \to 0} \frac{{\rm{x}}}{{{\rm{sinx}}}}\)

\(= 1 \cdot 1 = 1\)

Hence, option (3) is correct.

If\(\rm f(x)=\left\{\begin{matrix}\rm \dfrac{\sin 3x}{e^{2x}-1}, &\rm x \ne 0\\\rm k-2, &\rm x=0 \end{matrix}\right.\) is continuous at x = 0, then k = ?

1. \(\dfrac{3}{2}\)
2. \(\dfrac{9}{5}\)
3. \(\dfrac{1}{2}\)
4. \(\dfrac{7}{2}\)

Answer (Detailed Solution Below)

Option 4 : \(\dfrac{7}{2}\)

Concept:

Definition:

• A function f(x) is said to be continuous at a point x = a in its domain, if\(\rm \displaystyle \lim_{x\to a}f(x)\) exists oror if its graph is a single unbroken curve at that point.
• f(x) is continuous at x = a ⇔\(\rm \displaystyle \lim_{x\to a^+}f(x)=\lim_{x\to a^-}f(x)=\lim_{x\to a}f(x)=f(a)\).

Formulae:

• \(\rm \displaystyle \lim_{x\to 0}\dfrac{\sin x}{x}=1\)
• \(\rm \displaystyle \lim_{x\to 0}\dfrac{e^x-1}{x}=1\)

Calculation:

Since f(x) is given to be continuous at x = 0,\(\rm \displaystyle \lim_{x\to 0}f(x)=f(0)\).

Also,\(\rm \displaystyle \lim_{x\to a^+}f(x)=\lim_{x\to a^-}f(x)\) because f(x) is same for x > 0 and x < 0.

\(\rm \therefore \displaystyle \lim_{x\to 0}f(x)=f(0)\)

\(\rm \Rightarrow \displaystyle \lim_{x\to 0}\dfrac{\sin 3x}{e^{2x}-1}=k-2\)

\(\rm \Rightarrow \displaystyle \lim_{x\to 0}\dfrac{\dfrac{\sin 3x}{3x}\times3x}{\dfrac{e^{2x}-1}{2x}\times2x}=k-2\)

\(\rm \Rightarrow \dfrac{3}{2}=k-2\)

\(\rm \Rightarrow k=\dfrac{7}{2}\).

\(\mathop {\lim }\limits_{n \to \infty } \frac{{{2^{n + 1}} + {3^{n + 1}}}}{{{2^n} + {3^n}}}\) equals

1. 3
2. 2
3. 1
4. 0

Answer (Detailed Solution Below)

Option 1 : 3

\(\mathop {\lim }\limits_{n \to \infty } \frac{{{2^{n + 1}} + {3^{n + 1}}}}{{{2^n} + {3^n}}} \)

This can be written as:

\(= \mathop {\lim }\limits_{n \to \infty } \frac{{{2^n}2 + {3^n}3}}{{{2^n} + {3^n}}}\)

Taking 3ⁿ common, we can write:

\( = \mathop {\lim }\limits_{n \to \infty } \frac{{{3^n}\left[ {2.{{\left( {\frac{2}{3}} \right)}^n} + 3} \right]}}{{{3^n}\left[ {{{\left( {\frac{2}{3}} \right)}^n} + 1} \right]}}\)

\( = \mathop {\lim }\limits_{n \to \infty } \frac{{2.{{\left( {\frac{2}{3}} \right)}^n} + 3}}{{\left[ {{{\left( {\frac{2}{3}} \right)}^n} + 1} \right]}}\)

Here\(\frac 2 3 < 1\)

So,\(\left[ \frac {2}{3}\right]^{\infty} = 0\)

\(= \frac{{0 + 3}}{{0 + 1}} = 3\)

What is\(\rm \displaystyle\lim_{x \rightarrow 0} \dfrac{3^x + 3^{-x}-2}{x}\) equal to ?

1. 0
2. -1
3. 1
4. Limit does not exist

Answer (Detailed Solution Below)

Option 1 : 0

Concept:

\(\rm ​​\mathop {\lim }\limits_{x\; \to \;a} \left[ {f\left( x \right) + g\left( x \right)} \right] = \;\mathop {\lim }\limits_{x\; \to \;a} f\left( x \right) + \;\mathop {\lim }\limits_{x\; \to \;a} g\left( x \right)\)

\(\rm ​​\mathop {\lim }\limits_{x\; \to \;0} \dfrac { (a^x - 1) }{x} = \log a\)

log mⁿ = n log m

Calculation:

\(\rm \displaystyle\lim_{x \rightarrow 0} \dfrac{3^x + 3^{-x}-2}{x}\\= \displaystyle\lim_{x \rightarrow 0} \dfrac{3^x -1+ 3^{-x}-1}{x}\\= \displaystyle\lim_{x \rightarrow 0} \dfrac{3^x -1}{x}+\displaystyle\lim_{x \rightarrow 0} \dfrac{3^{-x} -1}{x}\\= \displaystyle\lim_{x \rightarrow 0} \dfrac{3^x -1}{x}+\displaystyle\lim_{x \rightarrow 0} \dfrac{(3^{-1})^x -1}{x}\\= \log 3 + \log (3^{-1})\\= \log 3 - \log 3\\=0\)

If\(\rm f(x) = \left\{ \begin{matrix} \rm \frac{2x-\sin^{-1} x}{2x + \tan^{-1} x}; & \rm x \ne 0 \\ \rm K; & \rm x = 0 \end{matrix}\right.\) is a continuous function at x = 0, then the value of k is:

1. 2
2. \(\dfrac12\)
3. 1
4. None of these

Answer (Detailed Solution Below)

Option 4 : None of these

Concept:

Definition:

• A function f(x) is said tobe continuous at a point x = a in its domain, if\(\rm \displaystyle \lim_{x\to a}f(x)\) existsoror if its graph is a single unbroken curve at that point.
• f(x) is continuous at x = a ⇔\(\rm \displaystyle \lim_{x\to a^+}f(x)=\lim_{x\to a^-}f(x)=\lim_{x\to a}f(x)=f(a)\).

Calculation:

For x ≠ 0, the given function can be re-written as:

\(\rm f(x) = \left\{ \begin{matrix} \rm \frac{2x-\sin^{-1} x}{2x + \tan^{-1} x}; & \rm x \ne 0 \\ \rm K; & \rm x = 0 \end{matrix}\right.\)

Since the equation of the function is same for x < 0 and x > 0, we have:

\(\rm \displaystyle \lim_{x\to 0^+}f(x)=\lim_{x\to 0^-}f(x)=\lim_{x\to 0}\frac{2x-\sin^{-1} x}{2x + \tan^{-1} x}\)

=\(\rm \displaystyle \lim_{x\to 0}\frac{2- \frac {\sin^{-1} x} x}{2 +\frac {\tan^{-1} x} x} = \frac {2-1}{2+1} = \frac 1 3\)

For the function to be continuous at x = 0, we must have:

\(\rm \displaystyle \lim_{x\to 0}f(x)=f(0)\)

⇒ K =\(\dfrac{1}{3}\).

The value of\(\rm \mathop {\lim }\limits_{x \to 0} \frac{{x}}{\cos x }\) is

1. 1
2. 0
3. ∞
4. -1

Answer (Detailed Solution Below)

Option 2 : 0

Calculation:

Here we have to find the value of\(\rm \mathop {\lim }\limits_{x \to 0} \frac{{x}}{\cos x }\)

Let L =\(\rm \mathop {\lim }\limits_{x \to 0} \frac{{x}}{\cos x }\)

Now put the value of limit, we get

=\(\rm \frac {0}{\cos 0} = \frac 01=0\)

What is\(\rm \displaystyle\lim_{x\rightarrow 0} \dfrac{\sin x \log (1-x)}{x^2}\) equal to?

1. -1
2. Zero
3. -e
4. \(\rm -\dfrac{1}{e}\)

Answer (Detailed Solution Below)

Option 1 : -1

Concept:

\(\rm ​​\mathop {\lim }\limits_{x\; \to \;a} \left[ {f\left( x \right) \cdot g\left( x \right)} \right] = \;\mathop {\lim }\limits_{x\; \to \;a} f\left( x \right) \cdot \;\mathop {\lim }\limits_{x\; \to \;a} g\left( x \right)\)

\(\rm ​​\mathop {\lim }\limits_{x\; \to \;0} \dfrac {\sin x }{x} = 1\\\rm ​​\mathop {\lim }\limits_{x\; \to \;0} \dfrac {\log (1+x) }{x}\)

Calculation:

We have to find the value of\(\rm \displaystyle\lim_{x\rightarrow 0} \dfrac{\sin x \log (1-x)}{x^2}\)

As we know,\(\rm ​​\mathop {\lim }\limits_{x\; \to \;a} \left[ {f\left( x \right) \cdot g\left( x \right)} \right] = \;\mathop {\lim }\limits_{x\; \to \;a} f\left( x \right) \cdot \;\mathop {\lim }\limits_{x\; \to \;a} g\left( x \right)\)

\(\therefore \rm \displaystyle\lim_{x\rightarrow 0} \dfrac{\sin x \log (1-x)}{x^2}= \rm ​​\mathop {\lim }\limits_{x\; \to \;0} \dfrac {\sin x }{x} × ​​\mathop {\lim }\limits_{x\; \to \;0} \dfrac {\log (1-x) }{x}\)

= 1 × \(\rm ​​\mathop {\lim }\limits_{x\; \to \;0} \dfrac {\log (1+(-x)) }{x}\)

= \(\rm ​​\mathop {\lim }\limits_{x\; \to \;0} \dfrac {\log (1+(-x)) }{-(-x)}\)

=\(-1 × \rm ​​\mathop {\lim }\limits_{x\; \to \;0} \dfrac {\log (1+(-x)) }{(-x)}\)

= -1 × 1

= -1

Let f(2) = 2 and f'(2) = 2. Then \(\mathop {\lim }\limits_{{\rm{x}} \to 2} \frac{{{\rm{xf}}\left( 2 \right) - 2{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{x}} - 2}}\) is given by

1. -4
2. 2
3. -2
4. 4

Answer (Detailed Solution Below)

Option 3 : -2

Concept:

• L' hospital's Rule: For (0/0), (∞ /∞) form of limits

If suppose,\(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = \frac{0}{0}{\rm{\;OR\;}}\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = \frac{{ \pm \infty }}{{ \pm \infty }}{\rm{\;\;}}\)

Then we apply L-Hospital's Rule,

\(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = \mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{\rm{f}}\left( {\rm{x}} \right)} \right)}}{{\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{\rm{g}}\left( {\rm{x}} \right)} \right)}} = \mathop {{\rm{lim}}}\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f'}}\left( {\rm{x}} \right)}}{{{\rm{g'}}\left( {\rm{x}} \right)}}{\rm{\;}}\)

Differentiate the numerator and differentiate the denominator and then take limit

Calculation:

Given: f(2) = 2 and f'(2) = 2.

To find: \(\mathop {\lim }\limits_{{\rm{x}} \to 2} \frac{{{\rm{xf}}\left( 2 \right) - 2{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{x}} - 2}}{\rm{\;}}\)=?

Check the form by putting x = 2

\(\mathop {\lim }\limits_{{\rm{x}} \to 2} \frac{{{\rm{xf}}\left( 2 \right) - 2{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{x}} - 2}} = \frac{{2\left( 2 \right) - 2\left( 2 \right)}}{{2 - 2}} = \frac{0}{0}\)

Apply L' hospital's rule,

\(\mathop {\lim }\limits_{{\rm{x}} \to 2} \frac{{{\rm{xf}}\left( 2 \right) - 2{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{x}} - 2}} = \mathop {\lim }\limits_{{\rm{x}} \to 2} \frac{{\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{\rm{xf}}\left( 2 \right) - 2{\rm{f}}\left( {\rm{x}} \right)} \right)}}{{\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{\rm{x}} - 2} \right)}} = \mathop {{\rm{lim}}}\limits_{{\rm{x}} \to 2} \frac{{{\rm{f}}\left( 2 \right) - 2{\rm{f'}}\left( {\rm{x}} \right)}}{1}\)

Differentiate the numerator and differentiate the denominator, we get

Now, take limit at x = 2

\(\mathop {{\rm{lim}}}\limits_{{\rm{x}} \to 2} = {\rm{f}}\left( 2 \right) - 2{\rm{f'}}\left( 2 \right)\)

\(= 2 - 2\left( 2 \right) = - 2\)

Option (3) is correct.

What is the value of\(\rm \mathop {{\rm{lim}}}\limits_{x \to 0} \;\;\frac{{{{\left( {1 - \cos2x} \right)}^3}\;}}{{{x^6}}}\)

1. 2
2. 4
3. 8
4. Limit doesn't exist

Answer (Detailed Solution Below)

Option 3 : 8

Concept:

Formulas:

1 - cos 2θ = 2 sin2 θ

1 + cos 2θ = 2 cos2 θ

\(\rm \mathop {{\rm{lim}}}\limits_{x \to 0} \;\;\frac{{\sin x}}{x} = 1\)

Calculation:

To Find: Value of\(\rm \mathop {{\rm{lim}}}\limits_{x \to 0} \;\;\frac{{{{\left( {1 - \cos2x} \right)}^3}\;}}{{{x^6}}}\)

\(\rm \mathop {{\rm{lim}}}\limits_{x \to 0} \;\;\frac{{{{\left( {1 - \cos2x} \right)}^3}\;}}{{{x^6}}}\)

=\(\rm \mathop {{\rm{lim}}}\limits_{x \to 0} \;\;\frac{{{{\left( {2{{\sin }^2}x} \right)}^3}}}{{{x^6}}}\)          (∵1 - cos 2θ = 2 sin2 θ)

=\(\rm \mathop {{\rm{lim}}}\limits_{x \to 0} \;\;\frac{{8{{\sin }^6}x}}{{{x^6}}}\)

=\(\rm \mathop {\lim }\limits_{x \to 0} 8\; \times \;{\left( {\frac{{\sin x}}{x}} \right)^6}\)

= 8 × 1

= 8

Evaluate\(\rm \displaystyle \lim_{x \rightarrow 0} \frac{1 - \cos 2x }{x^2} \)

1. 0
2. 1
3. 2
4. does not exists

Answer (Detailed Solution Below)

Option 3 : 2

Concept:

\(\rm \displaystyle \lim_{x \rightarrow 0} \frac{\sin x }{x} =1\)

Trigonometry formulas:

1 - cos 2x = 2sin² x

1 + cos 2x = 2cos² x

Calculation:

Here, we have to find the value of the limit\(\rm \displaystyle \lim_{x \rightarrow 0} \frac{1 - \cos 2x }{x^2} \)

As we know, 1 - cos 2x = 2sin2 x

\(\rm \displaystyle \lim_{x \rightarrow 0} \frac{1 - \cos 2x }{x^2} \)

=\(\rm \displaystyle \lim_{x \rightarrow 0} \frac{2\sin^2 x }{x^2} \)

= 2 ×\(\rm \displaystyle \lim_{x \rightarrow 0} \frac{\sin x }{x} × \lim_{x \rightarrow 0} \frac{\sin x }{x}\)

= 2 × 1 × 1

= 2

Function f(x) = tan(2x) will be discontinuous at x =

1. (2n + 1) π
2. (n + 1) π/2
3. (2n + 1) π/2
4. (2n + 1) π/4

Answer (Detailed Solution Below)

Option 4 : (2n + 1) π/4

Concept:

• f(c) must be defined. You can't have a hole in the function (denominator is zero i.e. infinity)
• cos x = 0, where x = odd multiple of π/2 = (2n + 1) π/2

Calculation:

Here, \(\tan \left( {2{\rm{x}}} \right) = \frac{{\sin \left( {2{\rm{x}}} \right)}}{{\cos \left( {2{\rm{x}}} \right)}}\)

Check where denominator become zero

We know cos x = 0, where x = (2n + 1) π/2

So for cos 2x = 0 where, 2x = (2n + 1) π/2

x = (2n + 1) π/4

∴ Given function is discontinuous at x = (2n + 1) π/4

Hence, option (4) is correct.

Hint:

• When dealing with a rational expression in which both the numerator and denominator are continuous.
• The only points in which the rational expression will be discontinuous where denominator become zero.

\(\mathop {\lim }\limits_{{\rm{x}} \to 2} \sqrt {{{\rm{x}}^2} + 5} = ?\)

1. √5
2. √7
3. √9
4. Not possible

Answer (Detailed Solution Below)

Option 3 : √9

Concept:

• \(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \sqrt[{\rm{n}}]{{{\rm{f}}\left( {\rm{x}} \right)}} = \sqrt[{\rm{n}}]{{\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} {\rm{f}}\left( {\rm{x}} \right)}}\)
• \(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} {\rm{f}}\left( {\rm{x}} \right) + {\rm{g}}\left( {\rm{x}} \right) = \mathop {{\rm{lim}}}\limits_{{\rm{x}} \to {\rm{a}}} {\rm{f}}\left( {\rm{x}} \right) + \mathop {{\rm{lim}}}\limits_{{\rm{x}} \to {\rm{a}}} {\rm{g}}\left( {\rm{x}} \right)\)

Calculation:

To find:\(\mathop {\lim }\limits_{{\rm{x}} \to 2} \sqrt {{{\rm{x}}^2} + 5} \) = ?

\(\mathop {\lim }\limits_{{\rm{x}} \to 2} \sqrt {{{\rm{x}}^2} + 5} = \sqrt {\mathop {\lim }\limits_{{\rm{x}} \to 2}( {{\rm{x}}^2} + 5}) \)

\(= \sqrt {\mathop {\lim }\limits_{{\rm{x}} \to 2} {{\rm{x}}^2} + \mathop {{\rm{lim}}}\limits_{{\rm{x}} \to 2} 5} = \sqrt {{2^2} + 5} \)

\(= \sqrt 9 \)

Hence, option (3) is correct.

Evaluate\(\mathop {\lim }\limits_{{\rm{x\;}} \to 0} \frac{{\rm \sqrt {a+x} - {\rm{\;}}\sqrt {\rm a - x} }}{\rm 4x}\)

1. \( \frac{1}{{2\sqrt 2 }}\)
2. \( \frac{1}{{\sqrt 2 }}\)
3. \(\frac{1}{2}\)
4. None of the above

Answer (Detailed Solution Below)

Option 4 : None of the above

Calculation:

We have to find the value of\(\mathop {\lim }\limits_{{\rm{x\;}} \to 0} \frac{{\rm \sqrt {a+x} - {\rm{\;}}\sqrt {\rm a - x} }}{\rm 4x}\)

Rationalise the numerator, we get

\(\mathop {\lim }\limits_{{\rm{x\;}} \to 0} \frac{{\rm \sqrt {a+x} - {\rm{\;}}\sqrt {\rm a - x} }}{\rm 4x} \times \frac{{\rm \sqrt {a+x} + {\rm{\;}}\sqrt {\rm a - x} }}{{\rm \sqrt {a+x} + {\rm{\;}}\sqrt {\rm a - x} }} \)

=\(\mathop {\lim }\limits_{{\rm{x\;}} \to 0} \frac{{\rm {a+x} - {\rm{\;}} {\rm (a - x )} }}{\rm 4x({\rm \sqrt {a+x} + {\rm{\;}}\sqrt {\rm a - x} } )} \)

=\(\mathop {\lim }\limits_{{\rm{x\;}} \to 0} \frac{\rm 2x}{\rm 4x({\rm \sqrt {a+x} + {\rm{\;}}\sqrt {\rm a - x} } )} \)

=\(\mathop {\lim }\limits_{{\rm{x\;}} \to 0} \frac{\rm 1}{\rm 2({\rm \sqrt {a+x} + {\rm{\;}}\sqrt {\rm a - x} } )} \)

=\(\frac{1}{{4 \rm \sqrt a }}\)

Find the value of\(\rm \displaystyle \lim_{n \rightarrow \infty} \frac{1-n^2} {\sum n}\)

1. -2
2. 2
3. 1
4. - 1

Answer (Detailed Solution Below)

Option 1 : -2

Concept:

\(\rm \sum n = \frac{n(n+1)}{2}\)

Calculation:

\(\rm \displaystyle \lim_{n \rightarrow \infty} \frac{1-n^2} {\sum n}\)

=\(\rm \displaystyle \lim_{n \rightarrow \infty} \frac{1-n^2} { \left[\frac{n(n+1)}{2} \right ]}\)

=\(\rm \displaystyle \lim_{n \rightarrow \infty} \frac{2(1-n^2)} {n^2+n}\)

Divide both numerator and denominator by n²

=\(\rm \displaystyle \lim_{n \rightarrow \infty} \frac{2\left[\frac{1-n^2}{n^2} \right ]} {\left[\frac{n^2+n}{n^2} \right ]}\)

=\(\rm \displaystyle \lim_{n \rightarrow \infty} \frac{2\left[\frac{1}{n^2} - 1 \right ]} {1 + \frac{1}{n} }\)

=\(\rm \frac{2(0-1)}{(1+0)}\)

= -2

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Source: https://testbook.com/objective-questions/mcq-on-limit-and-continuity--5eea6a1439140f30f369f1a7

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