Limit and Continuity MCQ Quiz - Objective Question with Answer for Limit and Continuity - Download Free PDF

Concept:

• Limit: \( \displaystyle\lim_{x\to0}f(x) \) exists if left-hand and right-hand limits coincide.
• Continuity: \( f \) is continuous at \( x=0 \) when \( \displaystyle\lim_{x\to0}f(x)=f(0) \).
• Differentiability: \( f \) is differentiable at \( x=0 \) if \( f'(0)=\displaystyle\lim_{h\to0}\frac{f(h)-f(0)}{h} \) exists.
• Derivative limit: Even if \( f'(0) \) exists, \( \displaystyle\lim_{x\to0}f'(x) \) may fail to exist due to oscillation.

Calculation:

Given,

\( f(x)=\begin{cases}x^{2}\sin\!\left(\dfrac{1}{x}\right),& x\ne0\\[4pt]0,& x=0\end{cases} \)

⇒ \( |x^{2}\sin(1/x)|\le x^{2} \)

⇒ \( \displaystyle\lim_{x\to0}f(x)=0 \)

⇒ limit exists

⇒ \( f(0)=0 \)

⇒ limit = value

⇒ continuous

⇒ \( f'(0)=\displaystyle\lim_{h\to0}\frac{h^{2}\sin(1/h)}{h} \)

⇒ \( =\lim_{h\to0}h\sin(1/h)=0 \)

⇒ differentiable at 0

⇒ For \( x\ne0 \), \( f'(x)=2x\sin(1/x)-\cos(1/x) \)

⇒ \( 2x\sin(1/x)\to0 \) as \( x\to0 \)

⇒ \( -\cos(1/x) \) oscillates

⇒ overall limit of \( f'(x) \) as \( x\to0 \) does not exist

∴ statements 1, 2, 3 and 4 are all true.

Calculation:

Given,

The function is \( f(x) = x^2 + 9 \).

Statement I: f(x) is an increasing function.

The derivative of f(x) is:

\( f'(x) = \frac{d}{dx}(x^2 + 9) = 2x \)

When \(( x > 0 )\), \(( f'(x) > 0 )\), so (f(x) is increasing.

When (x < 0), f'(x) < 0 ), so f(x)  is decreasing.

At (x = 0), ( f'(x) = 0 ), meaning the function is neither increasing nor decreasing at this point.

Hence, f(x)  is not entirely increasing. It is increasing for (x > 0) and decreasing for ( x < 0).

Statement II: f(x) has local maximum at x = 0

Since the function\( f(x) = x^2 + 9 \) is a parabola opening upwards (because the coefficient of x² is positive), it has a global minimum at x = 0, not a local maximum.

Conclusion:

- Statement I is incorrect because the function is not entirely increasing. It is increasing for x > 0  and decreasing for  x < 0 .

- Statement II is incorrect because the function has a global minimum at x = 0, not a local maximum.

Hence, the correct answer is Option 4. 

Calculation:

Given,

The function is \( f(x) = \sqrt{x^2 + 9} - 3 \) and \( g(x) = \sqrt{x^2 + 16} - 4 \).

We are tasked with finding:

\( \lim_{x \to 0} \frac{f(x)}{g(x)} \)

Multiply both the numerator and denominator by their respective conjugates:

\( \frac{\sqrt{x^2 + 9} - 3}{\sqrt{x^2 + 16} - 4} \times \frac{\sqrt{x^2 + 9} + 3}{\sqrt{x^2 + 9} + 3} \times \frac{\sqrt{x^2 + 16} + 4}{\sqrt{x^2 + 16} + 4} \)

Simplify the numerator:

\( (\sqrt{x^2 + 9} - 3)(\sqrt{x^2 + 9} + 3) = x^2 \)

Simplify the denominator:

\( (\sqrt{x^2 + 16} - 4)(\sqrt{x^2 + 16} + 4) = x^2 \)

Now, the expression becomes:

\( \frac{x^2}{x^2} \times \frac{\sqrt{x^2 + 16} + 4}{\sqrt{x^2 + 9} + 3} \)

Simplify and evaluate the limit:

\( \frac{\sqrt{x^2 + 16} + 4}{\sqrt{x^2 + 9} + 3} \) becomes:

\( \frac{\sqrt{16} + 4}{\sqrt{9} + 3} = \frac{4 + 4}{3 + 3} = \frac{8}{6} = \frac{4}{3} \)

Hence, the correct answer is Option 3.

Calculation:

Given,

The function is defined as:

\( f(x) = \begin{cases} x^3, & \text{if} \, |x| < 1 \\ x^2, & \text{if} \, |x| \geq 1 \end{cases} \)

We are tasked with finding:

\( \lim_{x \to -1} f'(x) \)

Check the left-hand limit for continuity at x = -1 :

\( \lim_{x \to -1^-} f(x) = (-1)^2 = 1 \)

Check the right-hand limit for continuity at x = -1 :

\( \lim_{x \to -1^+} f(x) = (-1)^3 = -1 \)

Since the left-hand limit (L.H.S) and right-hand limit (R.H.S) are not equal, the function is discontinuous at x = -1 .

Check the differentiability at x = 1 :

The left-hand derivative at x = 1  is \(\text{L.H.D} = 3 \) and the right-hand derivative at  x = 1  is \(\text{R.H.D} = 2 \) which means the function is not differentiable at x = 1 

∴ The function is neither continuous at x = -1  nor differentiable at x = 1 . 

Hence, the correct answer is Option 4.

Calculation:

Given,

The function is defined as:

\( f(x) = \begin{cases} x^3, & \text{if} \, |x| < 1 \\ x^2, & \text{if} \, |x| \geq 1 \end{cases} \)

We are tasked with finding:

\( \lim_{x \to 0} f'(x) \)

For  |x| < 1 , the function is  f(x) = x³, so the derivative is:

\( f'(x) = 3x^2 \)

Now, compute the limit of the derivative as x to 0:

\( \lim_{x \to 0} f'(x) = \lim_{x \to 0} 3x^2 = 0 \)

∴ The value of  \(\lim_{x \to 0} f'(x) \)is 0.

The correct answer is Option (c)

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 

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

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

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.

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

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

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.

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

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

Concept:

For a limit to exist, Left-hand limit and right-hand limit must be equal.

Calculations:

For a limit to exist Left-hand limit and right-hand limit must be equal.

|x| can have two values 

|x | = - x when x is negative 

|x| = x when x is positive.

\(\rm \displaystyle\lim_{x\rightarrow 0^-} \dfrac{|x|}{x}\) = \(\rm \displaystyle\lim_{x\rightarrow 0} \dfrac{-x}{x} = -1\)

​\(\rm \displaystyle\lim_{x\rightarrow 0^+} \dfrac{|x|}{x}\)​ = \(\rm \displaystyle\lim_{x\rightarrow 0} \dfrac{x}{x} = 1\)

Here, \(\rm \displaystyle\lim_{x\rightarrow 0^-} \dfrac{|x|}{x} \neq \rm \displaystyle\lim_{x\rightarrow 0^+}\dfrac{|x|}{x}\)

Hence, \(\rm \displaystyle\lim_{x\rightarrow 0} \dfrac{|x|}{x}\)does not exist. 

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 or or 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.

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

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

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

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

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

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

Concept:​

• \(\rm \displaystyle \lim_{x\to0}\dfrac{\tan x}{x}=1\).
• \(\rm \dfrac{d}{dx}\tan x=\sec^2x\).
• \(\rm \dfrac{d}{dx}\sec x=\tan x\sec x\).
• \(\rm \dfrac{d}{dx}\left[f(x)\times g(x)\right]=f(x)\dfrac{d}{dx}g(x)+g(x)\dfrac{d}{dx}f(x)\).

Indeterminate Forms: Any expression whose value cannot be defined, like \(\dfrac00\), \(\pm\dfrac{\infty}{\infty}\), 0⁰, ∞⁰ etc.

• For the indeterminate form \(\dfrac 0 0\), first try to rationalize by multiplying with the conjugate, or simplify by cancelling some terms in the numerator and denominator. Else, use the L'Hospital's rule.
• L'Hospital's Rule: For the differentiable functions f(x) and g(x), the \(\rm \displaystyle \lim_{x\to c} \dfrac{f(x)}{g(x)}\), if f(x) and g(x) are both 0 or ±∞ (i.e. an Indeterminate Form) is equal to the \(\rm \displaystyle \lim_{x\to c} \dfrac{f'(x)}{g'(x)}\) if it exists.

Calculation:

\(\rm \displaystyle\lim_{x\to 0}\dfrac{\tan x - x}{x^2 \tan x}=\dfrac00\) is an indeterminate form. Let us simplify and use the L'Hospital's Rule.

\(\rm \displaystyle\lim_{x\to 0}\dfrac{\tan x - x}{x^2 \tan x}=\lim_{x\to 0}\left[\dfrac{\tan x - x}{x^3}\times\dfrac{x}{\tan x}\right]\).

We know that \(\rm \displaystyle\lim_{x\to 0}\dfrac{x}{\tan x}=1\), but \(\rm \displaystyle \lim_{x\to 0}\dfrac{\tan x - x}{x^3}\) is still an indeterminate form, so we use L'Hospital's Rule:

\(\rm \displaystyle \rm \displaystyle \lim_{x\to 0}\dfrac{\tan x - x}{x^3}=\lim_{x\to 0}\dfrac{\sec^2 x - 1}{3x^2}\), which is still an indeterminate form, so we use L'Hospital's Rule again:

\(\rm \displaystyle \lim_{x\to 0}\dfrac{\sec^2 x - 1}{3x^2}= \lim_{x\to 0}\dfrac{2\sec x(\sec x\tan x)}{6x}=\lim_{x\to 0}\dfrac{\sec^2 x\tan x}{3x}\), which is still an indeterminate form, so we use L'Hospital's Rule again:

\(\rm \displaystyle \lim_{x\to 0}\dfrac{\sec^2 x\tan x}{3x}=\lim_{x\to 0}\dfrac{\sec^2 x\sec^2 x+\tan x[2\sec x(\sec x \tan x)]}{3}=\dfrac{1}{3}\).

∴ \(\rm \displaystyle\lim_{x\to 0}\dfrac{\tan x - x}{x^2 \tan x}=1\times\dfrac{1}{3}=\dfrac{1}{3}\).

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 or or 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 = \(\dfrac{1}{3}\).