lim _{x rightarrow 0} frac{4^x-9^x}{x(4^x+9^x | vedclass

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

Question

(A) Let $L = \lim _{x \rightarrow 0} \frac{4^x-9^x}{x(4^x+9^x)}$.Since the limit is of the form $\frac{0}{0}$ as $x \rightarrow 0$,we apply $L$-Hospit

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $L = \lim _{x \rightarrow 0} \frac{4^x-9^x}{x(4^x+9^x)}$.Since the limit is of the form $\frac{0}{0}$ as $x \rightarrow 0$,we apply $L$-Hospital's rule:$L = \lim _{x \rightarrow 0} \frac{\frac{d}{dx}(4^x-9^x)}{\frac{d}{dx}(x(4^x+9^x))}$$L = \lim _{x \rightarrow 0} \frac{4^x \log 4 - 9^x \log 9}{(4^x+9^x) + x(4^x \log 4 + 9^x \log 9)}$Substituting $x = 0$:$L = \frac{4^0 \log 4 - 9^0 \log 9}{(4^0+9^0) + 0(4^0 \log 4 + 9^0 \log 9)}$$L = \frac{\log 4 - \log 9}{1 + 1} = \frac{\log(4/9)}{2}$$L = \frac{1}{2} \log(\frac{2^2}{3^2}) = \frac{1}{2} \cdot 2 \log(\frac{2}{3}) = \log(\frac{2}{3})$.

$\lim _{x \rightarrow 0} \frac{4^x-9^x}{x(4^x+9^x)}$ is equal to

Similar Questions

$\mathop {\lim }\limits_{x \to - 2} \frac{{{{\sin }^{ - 1}}(x + 2)}}{{{x^2} + 2x}}$ is equal to

Let $f(x) = \int_0^x (t + \sin(1 - e^t)) dt, x \in R$. Then $\lim_{x \rightarrow 0} \frac{f(x)}{x^3}$ is equal to

$\lim _{x \rightarrow \frac{\pi}{2}} \left(\frac{2x-\pi}{\cos x}\right)$ is equal to

$\mathop {Limit}\limits_{x \to 0^+} \frac{1}{x\sqrt{x}} \left( a \tan^{-1} \frac{\sqrt{x}}{a} - b \tan^{-1} \frac{\sqrt{x}}{b} \right)$ has the value equal to

$\mathop {\lim }\limits_{x \to 0} \frac{{{a^x} - {b^x}}}{{{e^x} - 1}} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module