Question

please solve this please please

Answer 1

Question:-

Evaluate $\displaystyle\sf{\lim_{x\to0}\dfrac {(e^{2x}-1)(1-\cos (4x))}{x^3}.}$

Solution:-

The identities used:

$\displaystyle\sf{\lim_{x\to0}\dfrac {e^x-1}{x}=1}$

$\displaystyle\sf{1-\cos x=2\sin^2\left (\dfrac {x}{2}\right)}$

$\displaystyle\sf{\lim_{x\to0}\dfrac {\sin x}{x}=1}$

Well, as $\sf{x\longrightarrow0,\ 2x\longrightarrow0.}$

Then,

$\displaystyle\longrightarrow\sf{\lim_{x\to0}\dfrac {(e^{2x}-1)(1-\cos (4x))}{x^3}=\lim_{x\to0}\dfrac {e^{2x}-1}{x}\cdot\lim_{x\to0}\dfrac {2\sin^2(2x)}{x^2}}$

$\displaystyle\longrightarrow\sf{\lim_{x\to0}\dfrac {(e^{2x}-1)(1-\cos (4x))}{x^3}=2\lim_{2x\to0}\dfrac {e^{2x}-1}{2x}\cdot4\lim_{2x\to0}\dfrac {\sin (2x)}{2x}\cdot2\lim_{2x\to0}\dfrac {\sin (2x)}{2x}}$

$\displaystyle\longrightarrow\sf{\lim_{x\to0}\dfrac {(e^{2x}-1)(1-\cos (4x))}{x^3}=2\times4\times2}$

$\displaystyle\longrightarrow\sf{\underline {\underline {\lim_{x\to0}\dfrac {(e^{2x}-1)(1-\cos (4x))}{x^3}=16}}}$