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  • What is the value of <img alt="\lim _{x \rightarrow 0}\left(1+x\left(2^x-1\right)\right)^{1 / 1-\cos x}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Clim%20_%7Bx%20%5Crightarrow%200

What is the value of \lim _{x \rightarrow 0}\left(1+x\left(2^x-1\right)\right)^{1 / 1-\cos x}?

Option: 1

1


Option: 2

2


Option: 3

5


Option: 4

4


Answers (1)

best_answer

Given: \lim _{x \rightarrow 0}\left(1+x\left(2^x-1\right)\right)^{1 / 1-\cos x}

If \lim _{x \rightarrow a} f(x)=\lim _{x \rightarrow a} g(x)=0$, then $e^{\lim _{x \rightarrow a} g(x)}.
=e^{\lim _{x \rightarrow 0} \frac{x\left(2^x-1\right)}{1-\cos x}}

Use the trigonometric formula 1-\cos x=2 \sin ^2\left(\frac{x}{2}\right).

=e^{\lim _{x \rightarrow 0} \frac{x\left(e^x-1\right)}{2 \sin ^2\left(\frac{x}{2}\right)}}

Dividing the numerator and denominator by x^{2},the expression becomes

=e^{\lim _{x \rightarrow 0}\frac{x\left ( 2^{x}-1 \right )/x^{2}}{2\sin ^{2}\left ( x/2 \right )/\left ( x/2 \right )^{2}4}}
=e^{\frac{4}{2}\lim _{x \rightarrow 0}\frac{\left ( 2^{x}-1 \right )/x}{\left ( \sin \left ( x/2 \right ) /\left ( x/2 \right )\right )^{2}}}
=e

Applying limit formula \lim _{x \rightarrow 0} \frac{\sin x}{x}=1$ and $\lim _{x \rightarrow 0} \frac{a^x-1}{x}=\log a, the above expression becomes,

=e^{\frac{4 \log _e 2}{2(1)}}
=e^{2 \log _e 2}
=e^{\log _e 2^2}
=e^{\log _e 4}
=4

Therefore, \lim _{x \rightarrow 0}\left(1+x\left(2^x-1\right)\right)^{1 / 2 \sin ^2\left(\frac{x}{2}\right)}=4.
 

Posted by

Divya Prakash Singh

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