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How to evaluate \mathrm{ \lim _{x \rightarrow \infty}\left(\frac{x+3}{\sqrt{x^2-5 x}}\right)^{x^2 \sin (1 / x)}}

Option: 1

\mathrm{e^{\frac{11}{2}}}


Option: 2

0


Option: 3

1


Option: 4

e^{\frac{9}{2}}


Answers (1)

best_answer

\mathrm{\lim _{x \rightarrow \infty}\left(x^2 \sin \left(\frac{1}{x}\right) \ln \left(\frac{x+3}{\sqrt{x^2-5 x}}\right)\right)=\lim _{t \rightarrow 0}\left(\frac{1}{t^2} \sin (t) \ln \left(\frac{\frac{1}{t}+3}{\sqrt{\frac{1}{t^2}-\frac{5}{t}}}\right)\right)}

\mathrm{\text { Now } \sin (x) \approx x, x \rightarrow 0 \text { so: }}

                                       \mathrm{\approx \lim _{t \rightarrow 0}\left(\frac{1}{t} \ln \left(\frac{(3 t+1) \sqrt{-5 t+1}}{1-5 t}\right)\right)}

At this point  i used the rule of the de l'Hôpital so: 

                          \mathrm{\lim _{t \rightarrow 0}\left(\frac{1}{t} \ln \left(\frac{(3 t+1) \sqrt{-5 t+1}}{1-5 t}\right)\right)=\lim _{t \rightarrow 0}\left(\frac{\frac{-15 t+11}{2(-5 t+1)(3 t+1)}}{1}\right)=\frac{11}{2}}

So: 

                             \mathrm{\lim _{x \rightarrow \infty}\left(\left(\frac{x+3}{\sqrt{x^2-5 x}}\right)^{x^2 \sin \left(\frac{1}{x}\right)}\right)=e^{\frac{11}{2}}}

Posted by

SANGALDEEP SINGH

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