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Evaluate \mathrm{\lim _{x \rightarrow \infty}\left(2^{\sin \left(\frac{x^2+5}{x+5}\right)}-2^{\sin (x-5)}\right) }

Option: 1

0


Option: 2

\frac{1}{2}


Option: 3

1


Option: 4

\frac{3}{2}


Answers (1)

best_answer

First note that,\mathrm{2^{\sin \left(\frac{x^2+5}{x+5}\right)}-2^{\sin (x-5)}=2^{\sin (x-5)}\left[2^{\sin \left(\frac{x^2+5}{x+5}\right)-\sin (x-5)}-1\right] .}

Now, since  \mathrm{2^{\sin (x-5)}} is bounded (it varies between 2 and 1/2), for the limit to be zero,

\mathrm{\lim _{x \rightarrow \infty} 2^{\sin \left(\frac{x^2+5}{x+5}\right)-\sin (x-5)}-1}

must be 0, In other words,

\mathrm{\lim _{x \rightarrow \infty} 2^{\sin \left(\frac{x^2+5}{\pi+5}\right)-\sin (x-5)}=1}

Note that

\mathrm{\lim _{x \rightarrow \infty} 2^{\sin \left(\frac{x^2+5}{x+5}\right)-\sin (x-5)}= 2^{\lim _{x \rightarrow \infty} \sin \left(\frac{x^2+5}{x+5}\right)-\sin (x-5)} .}

Therefore, you must show that

\mathrm{\lim _{x \rightarrow \infty} \sin \left(\frac{x^2+5}{x+5}\right)-\sin (x-5)=0}

Observe that

\mathrm{\lim _{x \rightarrow \infty} \frac{x^2+5}{x+5}-(x-5)=\lim _{x \rightarrow \infty} \frac{x^2+5-\left(x^2-5\right)}{x-5}=\lim _{x \rightarrow \infty} \frac{10}{x+5} .}Therefore, the difference between \frac{x^2+5}{x+5} and x-5 approches zero, By the mean value theorem,

\mathrm{\left(\sin \left(\frac{x^2+5}{x+5}\right)-\sin (x-5)\right)=\cos (c)\left|\frac{x^2+5}{x+5}-(x-5)\right|}

 

Where c is between \frac{x^2+5}{x+5} and x-5 (since \mathrm{cos (x)} is the derived of \mathrm{sin (x))}. since \mathrm{cos (x))} is a bounded function,\mathrm{\begin{gathered} \lim _{x \rightarrow \infty}\left(\sin \left(\frac{x^2+5}{x+5}\right)-\sin (x-5)\right)=\lim _{x \rightarrow \infty} \cos (c)\left|\frac{x^2+5}{x+5}-(x-5)\right| \leq \\ \lim _{x \rightarrow \infty}\left|\frac{x^2+5}{x+5}-(x-5)\right|=0 . \end{gathered}}

This completes the computation

 

 

Posted by

Divya Prakash Singh

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