lim _{x rightarrow 0} frac{3^{sin x}-2^{tan x | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) We use the standard limit $\lim _{u \rightarrow 0} \frac{a^u - 1}{u} = \ln a$. The given expression is $\lim _{x \rightarrow 0} \frac{3^{\sin x}-2

Vedclass · Accepted Answer

$\log _e \frac{3}{2}$

Vedclass · Answer

(D) We use the standard limit $\lim _{u ightarrow 0} \frac{a^u - 1}{u} = \ln a$. The given expression is $\lim _{x ightarrow 0} \frac{3^{\sin x}-2^{	an x}}{\sin x}$. Dividing the numerator and denominator by $x$,we get: $\lim _{x ightarrow 0} \frac{\frac{3^{\sin x}-1}{x} - \frac{2^{	an x}-1}{x}}{\frac{\sin x}{x}}$. Since $\lim _{x ightarrow 0} \frac{\sin x}{x} = 1$ and $\lim _{x ightarrow 0} \frac{	an x}{x} = 1$,we have: $\lim _{x}$ ${ightarrow 0} \left( \frac{3^{\sin x}-1}{\sin x} \cdot \frac{\sin x}{x} - \frac{2^{	an x}-1}{	an x} \cdot \frac{	an x}{x} ight) / \frac{\sin x}{x}$. This simplifies to $\ln 3 - \ln 2 = \ln \left( \frac{3}{2} ight)$.

$\lim _{x \rightarrow 0} \frac{3^{\sin x}-2^{\tan x}}{\sin x}=$

Similar Questions

The value of $\mathop {\lim }\limits_{x \to \infty } \left( {\frac{{{x^2} + bx + 4}}{{{x^2} + ax + 5}}} \right)$ is

If $a = \lim_{x \rightarrow 0} \frac{\sqrt{1+\sqrt{1+x^4}}-\sqrt{2}}{x^4}$ and $b = \lim_{x \rightarrow 0} \frac{\sin^2 x}{\sqrt{2}-\sqrt{1+\cos x}}$,then the value of $ab^3$ is

$\lim _{x \rightarrow 1} \left( \lim _{y \rightarrow \infty} y \left( (e^x)^{1/y} - 1 \right) \right) = $

$\lim _{x \rightarrow 0} \frac{x^4+x^3+x^2}{\sin ^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right) \cdot \tan ^{-1} x} = $

Evaluate the limit: $\mathop {\lim }\limits_{x \to 3} (x+3)$

Vedclass Test Series

Exam Paper Generator

Online Exam Module