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

Concept:

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

Calculation:

Given: f(x) = $\{\begin{array}{l}\mathrm{m}\mathrm{x}+1,\text{ if }\mathrm{x}\le \frac{\pi }{2}\\ \sin \mathrm{x}+\mathrm{n},\text{ if }\mathrm{x}>\frac{\pi }{2}\end{array}$

f($\frac{\pi }{2}$) = m × $\frac{\pi }{2}$ + 1

Left-hand limit = $\underset{h\to 0}{lim}f(\frac{\pi }{2}-h)$

$=\underset{h\to 0}{lim}m\times (\frac{\pi }{2}-h)+1$

Applying the limits:

Left- hand limit = m × $\frac{\pi }{2}$ + 1

Right-hand limit = $\underset{h\to 0}{lim}f(\frac{\pi }{2}+h)$

$=\underset{h\to 0}{lim}\sin (\frac{\pi }{2}+h)+n$

Applying the limits:

 $=\sin \frac{\pi }{2}+n$

Right-hand limit = 1 + n

For the function to be continuous at x = $\frac{\pi }{2}$,

Left-hand limit = Right-hand limit = f(π/2)

⇒ m× $\frac{\pi }{2}$ + 1 = 1 + n

⇒ n = $\frac{m\pi }{2}$

The correct answer is n = $\frac{m\pi }{2}$ .

Concept:

If a function is continuous at a point a, then

${\displaystyle \underset{\mathrm{x}\to \mathrm{a}}{lim}\mathrm{f}(\mathrm{x})=\mathrm{f}(\mathrm{a})}$

sin(∞) = a, Where -1≤ a ≤ 1

Calculation:

Given:

f(0) = 1

f(x) = x sin (1/x)

Checking continuity at x = 0

${\displaystyle \underset{\mathrm{x}\to 0}{lim}\mathrm{f}(\mathrm{x})=\mathrm{f}(0)}$

L.H.L

= ${\displaystyle \underset{\mathrm{x}\to 0}{lim}\mathrm{f}(\mathrm{x})={\displaystyle \underset{\mathrm{x}\to 0}{lim}\mathrm{x}\sin \left(\frac{1}{\mathrm{x}}\right)=0\times \sin \left(\frac{1}{0}\right)}}$

= 0 × sin(∞)

= 0 

R.H.L

= f(0) = 1

L.H.L ≠ R.H.L

Hence, function is discontinuous at x = 0.

Concept:

If a function is continuous at x = a, then L.H.L = R.H.L = f(a).

Left hand limit (L.H.L) of f(x) at x = a is $\underset{h\to 0}{lim}f(a-h)$ 

Right hand limit (R.H.L) of f(x) at x = a is $\underset{h\to 0}{lim}f(a+h)$

Calculation:

Concept:

The function f(x) is continuous at x = a if

 f(a-) = f(a) = f(a+)

Calculation:

Given, f(x) = |x| 

For x ≥ 0, f(x) = x

and for x < 0, f(x) = - x 

So function is continuous for x > 0 and x < 0

At x = 0, 

f(0-) = f(0) = f(0+) = 0

⇒ f(x) is continuous at x = 0

∴ The correct answer is option (1).

Concept:

The function f(x) is continuous at x = a if

 f(a-) = f(a) = f(a+)

Calculation:

Given, f(x) = |x| 

For x ≥ 0, f(x) = x

and for x < 0, f(x) = - x 

So function is continuous for x > 0 and x < 0

At x = 0, 

f(0-) = f(0) = f(0+) = 0

⇒ f(x) is continuous at x = 0

∴ The correct answer is option (1).

Concept:

${\displaystyle \underset{\mathrm{x}\to 0}{lim}\frac{\sin \mathrm{x}}{\mathrm{x}}=1}$

Calculation:

${\displaystyle \underset{\mathrm{x}\to \mathrm{\infty }}{lim}2\mathrm{x}\sin \left(\frac{4}{\mathrm{x}}\right)}$

= $2\times {\displaystyle \underset{\mathrm{x}\to \infty }{lim}\frac{\sin \left(\frac{4}{\mathrm{x}}\right)}{\left(\frac{1}{\mathrm{x}}\right)}}$

= $2\times {\displaystyle \underset{\mathrm{x}\to \infty }{lim}\frac{\sin \left(\frac{4}{\mathrm{x}}\right)}{\left(\frac{4}{\mathrm{x}}\right)}\times 4}$

Let $\frac{4}{\mathrm{x}}=\mathrm{t}$

If x → ∞ then t → 0

= $8\times {\displaystyle \underset{\mathrm{t}\to 0}{lim}\frac{\sin \mathrm{t}}{\mathrm{t}}}$

= 8 × 1 

= 8 

Concept:

• 1 - cos 2θ = 2 sin2 θ
• $\underset{x\to 0}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{\sin x}{x}=1$

Calculation:

$\underset{x\to 0}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{{\left(1-\cos 2x\right)}^2}{x^4}$

= $\underset{x\to 0}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{{\left(2{\sin }^{2}x\right)}^2}{x^4}$          (1 - cos 2θ = 2 sin² θ)

= $\underset{x\to 0}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{4{\sin }^{4}x}{x^4}$

= $\underset{x\to 0}{lim}4\times {\left(\frac{\sin x}{x}\right)}^4$

= 4 × 1 = 4

Concept:

$\underset{\mathrm{x}\to \mathrm{a}}{lim}\left[\frac{\mathrm{f}\left(\mathrm{x}\right)}{\mathrm{g}\left(\mathrm{x}\right)}\right]=\frac{\underset{\mathrm{x}\to \mathrm{a}}{lim}\mathrm{f}\left(\mathrm{x}\right)}{\underset{\mathrm{x}\to \mathrm{a}}{lim}\mathrm{g}\left(\mathrm{x}\right)},\mathrm{p}\mathrm{r}\mathrm{o}\mathrm{v}\mathrm{i}\mathrm{d}\mathrm{e}\mathrm{d}\underset{\mathrm{x}\to \mathrm{a}}{lim}\mathrm{g}\left(\mathrm{x}\right)\ne 0$

$\underset{\mathrm{x}\to 0}{lim}\frac{\tan \mathrm{x}}{\mathrm{x}}=1$

$\underset{\mathrm{x}\to 0}{lim}\frac{\log (1+\mathrm{x})}{\mathrm{x}}=1$

Calculation:

$\underset{\mathrm{x}\to 0}{lim}\frac{\log (1+2\mathrm{x})}{\tan 2\mathrm{x}}$

$$\begin{gathered} =\underset{\mathrm{x}\to 0}{lim}\frac{\frac{\log (1+2\mathrm{x})}{2\mathrm{x}}\times 2\mathrm{x}}{\frac{\tan 2\mathrm{x}}{2\mathrm{x}}\times 2\mathrm{x}} \\ =\frac{\underset{\mathrm{x}\to 0}{lim}\frac{\log (1+2\mathrm{x})}{2\mathrm{x}}}{\underset{\mathrm{x}\to 0}{lim}\frac{\tan 2\mathrm{x}}{2\mathrm{x}}} \end{gathered}$$

As we know $\underset{\mathrm{x}\to 0}{lim}\frac{\tan \mathrm{x}}{\mathrm{x}}=1$ and $\underset{\mathrm{x}\to 0}{lim}\frac{\log (1+\mathrm{x})}{\mathrm{x}}=1$

Therefore, $\underset{\mathrm{x}\to 0}{lim}\frac{\tan 2\mathrm{x}}{2\mathrm{x}}=1$ and $\underset{\mathrm{x}\to 0}{lim}\frac{\log (1+2\mathrm{x})}{2\mathrm{x}}=1$

Hence $\underset{\mathrm{x}\to 0}{lim}\frac{\log (1+2\mathrm{x})}{\tan 2\mathrm{x}}=\frac{1}{1}=1$

Calculation:

We have to find the value of $\underset{\mathrm{x}\to \infty }{lim}\frac{\mathrm{x}}{\sqrt{1+2{\mathrm{x}}^2}}$

$\underset{\mathrm{x}\to \infty }{lim}\frac{\mathrm{x}}{\sqrt{1+2{\mathrm{x}}^2}}$       [Form $\frac{\infty }{\infty }$]

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

$\underset{\mathrm{x}\to \infty }{lim}\frac{\mathrm{x}}{\sqrt{1+2{\mathrm{x}}^2}}$

= $\underset{\mathrm{x}\to \mathrm{\infty }}{lim}\frac{\mathrm{x}}{\mathrm{x}\sqrt{\frac{1}{{\mathrm{x}}^2}+2}}$

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

= $\underset{\mathrm{x}\to \mathrm{\infty }}{lim}\frac{1}{\sqrt{\frac{1}{{\mathrm{x}}^2}+2}}$

= $\frac{1}{\sqrt{\frac{1}{{\mathrm{\infty }}^2}+2}}=\frac{1}{\sqrt{0+2}}=\frac{1}{\sqrt{2}}$

Calculation:

We have to find the value of $\underset{\mathrm{x}\to \mathrm{\infty }}{lim}\frac{{\mathrm{x}}^2}{1+{\mathrm{x}}^2}$

$\underset{\mathrm{x}\to \mathrm{\infty }}{lim}\frac{{\mathrm{x}}^2}{1+{\mathrm{x}}^2}$       [Form $\frac{\infty }{\infty }$]

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

$\underset{\mathrm{x}\to \mathrm{\infty }}{lim}\frac{{\mathrm{x}}^2}{1+{\mathrm{x}}^2}$

= $\underset{\mathrm{x}\to \mathrm{\infty }}{lim}\frac{{\mathrm{x}}^2}{{\mathrm{x}}^2\left(\frac{1}{{\mathrm{x}}^2}+1\right)}$

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

= $\underset{\mathrm{x}\to \mathrm{\infty }}{lim}\frac{1}{\left(\frac{1}{{\mathrm{x}}^2}+1\right)}$

= $\frac{1}{\frac{1}{{\mathrm{\infty }}^2}+1}=\frac{1}{0+1}=1$

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.

${\displaystyle \underset{\mathrm{x}\to 0^{-}}{lim}{\displaystyle \frac{|\mathrm{x}|}{\mathrm{x}}}}$ = ${\displaystyle \underset{\mathrm{x}\to 0}{lim}{\displaystyle \frac{-\mathrm{x}}{\mathrm{x}}}=-1}$

​${\displaystyle \underset{\mathrm{x}\to 0^{+}}{lim}{\displaystyle \frac{|\mathrm{x}|}{\mathrm{x}}}}$​ = ${\displaystyle \underset{\mathrm{x}\to 0}{lim}{\displaystyle \frac{\mathrm{x}}{\mathrm{x}}}=1}$

Here, ${\displaystyle \underset{\mathrm{x}\to 0^{-}}{lim}{\displaystyle \frac{|\mathrm{x}|}{\mathrm{x}}}\ne {\displaystyle \underset{\mathrm{x}\to 0^{+}}{lim}{\displaystyle \frac{|\mathrm{x}|}{\mathrm{x}}}}}$

Hence, ${\displaystyle \underset{\mathrm{x}\to 0}{lim}{\displaystyle \frac{|\mathrm{x}|}{\mathrm{x}}}}$does not exist. 

Concept:

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

LHL = RHL = value of f(c)

i.e., $\underset{\mathrm{x}\to {\mathrm{c}}^{-}}{lim}\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{x}\to {\mathrm{c}}^{+}}{lim}\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{c}\right)$

Calculation:

$\underset{\mathrm{x}\to \mathrm{a}}{lim}\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{x}\to \mathrm{a}}{\mathrm{l}\mathrm{i}\mathrm{m}}\left(\mathrm{x}-2\right)\left(\mathrm{x}-3\right)$            (a ϵ Real numbers)

$=\underset{\mathrm{x}\to \mathrm{a}}{\mathrm{l}\mathrm{i}\mathrm{m}}{\mathrm{x}}^2-3\mathrm{x}-2\mathrm{x}+6$

$=\underset{\mathrm{x}\to \mathrm{a}}{\mathrm{l}\mathrm{i}\mathrm{m}}{\mathrm{x}}^2-5\mathrm{x}+6$

$={\mathrm{a}}^2-5\mathrm{a}+6$

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

Important tip:

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

Concept:

Definition:

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

Formulae:

• ${\displaystyle \underset{\mathrm{x}\to 0}{lim}{\displaystyle \frac{\sin \mathrm{x}}{\mathrm{x}}}=1}$
• ${\displaystyle \underset{\mathrm{x}\to 0}{lim}{\displaystyle \frac{{\mathrm{e}}^{\mathrm{x}}-1}{\mathrm{x}}}=1}$

Calculation: 

Since f(x) is given to be continuous at x = 0, ${\displaystyle \underset{\mathrm{x}\to 0}{lim}\mathrm{f}(\mathrm{x})=\mathrm{f}(0)}$.

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

 $\therefore {\displaystyle \underset{\mathrm{x}\to 0}{lim}\mathrm{f}(\mathrm{x})=\mathrm{f}(0)}$

$\Rightarrow {\displaystyle \underset{\mathrm{x}\to 0}{lim}{\displaystyle \frac{\sin 3\mathrm{x}}{{\mathrm{e}}^{2\mathrm{x}}-1}}=\mathrm{k}-2}$

$\Rightarrow {\displaystyle \underset{\mathrm{x}\to 0}{lim}{\displaystyle \frac{{\displaystyle \frac{\sin 3\mathrm{x}}{3\mathrm{x}}}\times 3\mathrm{x}}{{\displaystyle \frac{{\mathrm{e}}^{2\mathrm{x}}-1}{2\mathrm{x}}}\times 2\mathrm{x}}}=\mathrm{k}-2}$

$\Rightarrow {\displaystyle \frac{3}{2}}=\mathrm{k}-2$

$\Rightarrow \mathrm{k}={\displaystyle \frac{7}{2}}$.

Concept:​

• ${\displaystyle \underset{\mathrm{x}\to 0}{lim}{\displaystyle \frac{\tan \mathrm{x}}{\mathrm{x}}}=1}$.
• ${\displaystyle \frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}}\tan \mathrm{x}={\sec }^2\mathrm{x}$.
• ${\displaystyle \frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}}\sec \mathrm{x}=\tan \mathrm{x}\sec \mathrm{x}$.
• ${\displaystyle \frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}}[\mathrm{f}(\mathrm{x})\times \mathrm{g}(\mathrm{x})]=\mathrm{f}(\mathrm{x}){\displaystyle \frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}}\mathrm{g}(\mathrm{x})+\mathrm{g}(\mathrm{x}){\displaystyle \frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}}\mathrm{f}(\mathrm{x})$.

Indeterminate Forms: Any expression whose value cannot be defined, like ${\displaystyle \frac{0}{0}}$, $\pm {\displaystyle \frac{\mathrm{\infty }}{\mathrm{\infty }}}$, 0⁰, ∞⁰ etc.

• For the indeterminate form ${\displaystyle \frac{0}{0}}$, 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 ${\displaystyle \underset{\mathrm{x}\to \mathrm{c}}{lim}{\displaystyle \frac{\mathrm{f}(\mathrm{x})}{\mathrm{g}(\mathrm{x})}}}$, if f(x) and g(x) are both 0 or ±∞ (i.e. an Indeterminate Form) is equal to the ${\displaystyle \underset{\mathrm{x}\to \mathrm{c}}{lim}{\displaystyle \frac{{\mathrm{f}}^{\prime }(\mathrm{x})}{{\mathrm{g}}^{\prime }(\mathrm{x})}}}$ if it exists.

Calculation:

${\displaystyle \underset{\mathrm{x}\to 0}{lim}{\displaystyle \frac{\tan \mathrm{x}-\mathrm{x}}{{\mathrm{x}}^2\tan \mathrm{x}}}={\displaystyle \frac{0}{0}}}$ is an indeterminate form. Let us simplify and use the L'Hospital's Rule.

${\displaystyle \underset{\mathrm{x}\to 0}{lim}{\displaystyle \frac{\tan \mathrm{x}-\mathrm{x}}{{\mathrm{x}}^2\tan \mathrm{x}}}=\underset{\mathrm{x}\to 0}{lim}[{\displaystyle \frac{\tan \mathrm{x}-\mathrm{x}}{{\mathrm{x}}^3}}\times {\displaystyle \frac{\mathrm{x}}{\tan \mathrm{x}}}]}$.

We know that ${\displaystyle \underset{\mathrm{x}\to 0}{lim}{\displaystyle \frac{\mathrm{x}}{\tan \mathrm{x}}}=1}$, but ${\displaystyle \underset{\mathrm{x}\to 0}{lim}{\displaystyle \frac{\tan \mathrm{x}-\mathrm{x}}{{\mathrm{x}}^3}}}$ is still an indeterminate form, so we use L'Hospital's Rule:

${\displaystyle {\displaystyle \underset{\mathrm{x}\to 0}{lim}{\displaystyle \frac{\tan \mathrm{x}-\mathrm{x}}{{\mathrm{x}}^3}}=\underset{\mathrm{x}\to 0}{lim}{\displaystyle \frac{{\sec }^2\mathrm{x}-1}{3{\mathrm{x}}^2}}}}$, which is still an indeterminate form, so we use L'Hospital's Rule again:

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

${\displaystyle \underset{\mathrm{x}\to 0}{lim}{\displaystyle \frac{{\sec }^2\mathrm{x}\tan \mathrm{x}}{3\mathrm{x}}}=\underset{\mathrm{x}\to 0}{lim}{\displaystyle \frac{{\sec }^2\mathrm{x}{\sec }^2\mathrm{x}+\tan \mathrm{x}[2\sec \mathrm{x}(\sec \mathrm{x}\tan \mathrm{x})]}{3}}={\displaystyle \frac{1}{3}}}$.

∴ ${\displaystyle \underset{\mathrm{x}\to 0}{lim}{\displaystyle \frac{\tan \mathrm{x}-\mathrm{x}}{{\mathrm{x}}^2\tan \mathrm{x}}}=1\times {\displaystyle \frac{1}{3}}={\displaystyle \frac{1}{3}}}$.

Concept:

Definition:

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

Calculation:

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

$\mathrm{f}(\mathrm{x})=\{\begin{array}{cc}\frac{2\mathrm{x}-{\sin }^{-1}\mathrm{x}}{2\mathrm{x}+{\tan }^{-1}\mathrm{x}};& \mathrm{x}\ne 0\\ \mathrm{K};& \mathrm{x}=0\end{array}$

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

${\displaystyle \underset{\mathrm{x}\to 0^{+}}{lim}\mathrm{f}(\mathrm{x})=\underset{\mathrm{x}\to 0^{-}}{lim}\mathrm{f}(\mathrm{x})=\underset{\mathrm{x}\to 0}{lim}\frac{2\mathrm{x}-{\sin }^{-1}\mathrm{x}}{2\mathrm{x}+{\tan }^{-1}\mathrm{x}}}$

= ${\displaystyle \underset{\mathrm{x}\to 0}{lim}\frac{2-\frac{{\sin }^{-1}\mathrm{x}}{\mathrm{x}}}{2+\frac{{\tan }^{-1}\mathrm{x}}{\mathrm{x}}}=\frac{2-1}{2+1}=\frac{1}{3}}$

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

${\displaystyle \underset{\mathrm{x}\to 0}{lim}\mathrm{f}(\mathrm{x})=\mathrm{f}(0)}$

⇒ K = ${\displaystyle \frac{1}{3}}$.