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  • Computing:    <img alt="\lim _{x \rightarrow \infty} \frac{\sqrt{1-\cos ^2 \frac{1}{x}}\left(3^{\frac{1}{x}}-5^{\frac{-1}{x}}\right)}{\log _2\left(1+x^{-2}+x^{-3}\right)}" src="https

Computing:    \lim _{x \rightarrow \infty} \frac{\sqrt{1-\cos ^2 \frac{1}{x}}\left(3^{\frac{1}{x}}-5^{\frac{-1}{x}}\right)}{\log _2\left(1+x^{-2}+x^{-3}\right)}

Option: 1

(\log 15)(\log 2)


Option: 2

(\log 5)(\log 2)


Option: 3

0


Option: 4

1


Answers (1)

best_answer

Your idea is very good; the limit you get is forand you get t \rightarrow 0^{+}, so \sqrt{1-\cos ^2 t}=\sin t 

\lim _{t \rightarrow 0^{+}} \frac{\sin t\left(3^t-5^{-t}\right)}{\log _2\left(1+t^2+t^3\right)}=\lim _{t \rightarrow 0^{+}} \frac{\sin t}{t} \frac{3^t-5^{-t}}{t} \frac{t^2 \log 2}{\log \left(1+t^2+t^3\right)}

(where "log" denotes the natural logarithm) and you can compute separately the limit of the three factors. The first is known to be 1 . Then

\lim _{t \rightarrow 0^{+}} \frac{3^t-5^{-t}}{t}=\log 3+\log 5

because it's the derivative at0 \: \: of \: \: f(t)=3^t-5^{-t}. Alternatively, write it as

\lim _{t \rightarrow 0^{+}}\left(\frac{3^t-1}{t}+\frac{5^t-1}{t} \frac{1}{5^t}\right)
and use the fundamental limits (which is basically what you did).
For the last one, apply l'Hôpital (or Taylor):

\lim _{t \rightarrow 0^{+}} \frac{2 t \log 2}{\frac{2 t+3 t^2}{1+t^2+t^3}}=\lim _{t \rightarrow 0^{+}} \frac{2\left(1+t^2+t^3\right) \log 2}{2+3 t}=\log 2

So finally you get (\log 3+\log 5) \log 2=(\log 15)(\log 2)

Posted by

Ritika Harsh

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