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

Calculation:

Given,

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

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

The derivative of f(x) is:

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

When $(x>0)$, $(f^{\prime }(x)>0)$, 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:

$\underset{x\to 0}{lim}\frac{f(x)}{g(x)}$

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:

$f(x)=\{\begin{array}{ll}{x}^3,& \text{if}|x|<1\\ x^2,& \text{if}|x|\ge 1\end{array}$

We are tasked with finding:

$\underset{x\to -1}{lim}{f}^{\prime }(x)$

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

$\underset{x\to -1^{-}}{lim}f(x)=(-1{)}^2=1$

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

$\underset{x\to -1^{+}}{lim}f(x)=(-1{)}^3=-1$

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:

$f(x)=\{\begin{array}{ll}{x}^3,& \text{if}|x|<1\\ x^2,& \text{if}|x|\ge 1\end{array}$

We are tasked with finding:

$\underset{x\to 0}{lim}{f}^{\prime }(x)$

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

$f^{\prime }(x)=3x^2$

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

$\underset{x\to 0}{lim}{f}^{\prime }(x)=\underset{x\to 0}{lim}3x^2=0$

∴ The value of  $\underset{x\to 0}{lim}{f}^{\prime }(x)$is 0.

The correct answer is Option (c)

Calculation:

${\displaystyle \Rightarrow \underset{\mathrm{x}\to \frac{{\pi }^{+}}{2}}{lim}{\mathrm{e}}^{\frac{\cot 6\mathrm{x}}{\cot 4\mathrm{x}}}=\underset{\mathrm{x}\to \frac{{\pi }^{+}}{2}}{lim}{\mathrm{e}}^{\frac{\sin 4\mathrm{x}\cdot \cos 6\mathrm{x}}{\sin 6\mathrm{x}\cdot \cos 4\mathrm{x}}}={\mathrm{e}}^{2/3}}$

${\displaystyle \Rightarrow \underset{\mathrm{x}\to \frac{{\pi }^{-}}{2}}{lim}(1+|\cos \mathrm{x}|{)}^{\frac{\lambda }{\cos \mathrm{x}}|}={\mathrm{e}}^{\lambda }}$

⇒ f(π/2) = µ 

For continuous function ⇒ e^2/3 = e^λ = µ 

$\lambda =\frac{2}{3},\mu ={\mathrm{e}}^{2/3}$

Now, 6λ + 6log_eµ + µ⁶ – e^6λ = 10

Hence, the correct answer is Option 4. 

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:

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:

• 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:​

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

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 = ${\displaystyle \frac{1}{3}}$.