lim _{x rightarrow 0} frac{(3^{2x}-sqrt{x+1}) | vedclass

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

Question

(B) Let $L = \lim _{x \rightarrow 0} \frac{(3^{2x} - \sqrt{x+1}) \sin 5x}{1 - \cos 4x}$.Using the identity $1 - \cos 4x = 2 \sin^2 2x$,we have $L = \l

Vedclass · Accepted Answer

$\frac{5}{16} \log \left(\frac{81}{e}\right)$

Vedclass · Answer

(B) Let $L = \lim _{x ightarrow 0} \frac{(3^{2x} - \sqrt{x+1}) \sin 5x}{1 - \cos 4x}$.Using the identity $1 - \cos 4x = 2 \sin^2 2x$,we have $L = \lim _{x ightarrow 0} \frac{(3^{2x} - \sqrt{x+1}) \sin 5x}{2 \sin^2 2x}$.Divide numerator and denominator by $x^2$ and use standard limits $\lim_{x 	o 0} \frac{\sin ax}{ax} = 1$:$L = \lim _{x ightarrow 0} \left( \frac{3^{2x} - \sqrt{x+1}}{x} ight) \cdot \left( \frac{\sin 5x}{x} ight) \cdot \left( \frac{x^2}{2 \sin^2 2x} ight) = \lim _{x ightarrow 0} \left( \frac{3^{2x} - 1 - (\sqrt{x+1} - 1)}{x} ight) \cdot 5 \cdot \frac{1}{2(2)^2}$.$L = \frac{5}{8} \lim _{x ightarrow 0} \left( \frac{3^{2x} - 1}{x} - \frac{\sqrt{x+1} - 1}{x} ight)$.Using $\lim_{x 	o 0} \frac{a^x - 1}{x} = \ln a$ and $\lim_{x 	o 0} \frac{\sqrt{x+1} - 1}{x} = \frac{1}{2}$:$L = \frac{5}{8} \left( 2 \ln 3 - \frac{1}{2} ight) = \frac{5}{16} (4 \ln 3 - 1) = \frac{5}{16} (\ln 81 - \ln e) = \frac{5}{16} \ln \left( \frac{81}{e} ight)$.

$\lim _{x \rightarrow 0} \frac{(3^{2x}-\sqrt{x+1}) \sin 5x}{1-\cos 4x} =$

Similar Questions

Evaluate the given limit: $\mathop {\lim }\limits_{x \to 1} \frac{a x^{2}+b x+c}{c x^{2}+b x+a}$,where $a+b+c \neq 0$.

$\lim _{x \rightarrow 2}\left(\frac{5 x-8}{8-3 x}\right)^{\frac{3}{2 x-4}} = $

If $f(x) = \begin{cases} \frac{\sin([x])}{[x]}, & \text{when } [x] \neq 0 \\ 0, & \text{when } [x] = 0 \end{cases}$ where $[x]$ is the greatest integer function,then $\lim_{x \to 0} f(x) = $

$\mathop {\lim }\limits_{x \to \infty } \frac{{\sqrt {{x^2} + {a^2}} - \sqrt {{x^2} + {b^2}} }}{{\sqrt {{x^2} + {c^2}} - \sqrt {{x^2} + {d^2}} }} = $

$\lim _{x \rightarrow 3} \frac{x^3-27}{x^2-9} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module