Solveeit Logo
Login

Question

Mathematics Question on limits and derivatives

lim⁡x→∞(x100ex+(cos2x)x2)=\displaystyle \lim_{x\to\infty}\left(\frac{x^{100}}{e^{x}}+\left(cos \frac{2}{x}\right)^{x^2}\right) =

A

e−1e^{-1}

B

e−4e^{-4}

C

(1+e−2)(1+e^{-2})

D

e−2e^{-2}

Answer

e−2e^{-2}

Explanation

Solution

Consider lim⁡x→∞[x100ex+(cos2x)x2]\displaystyle \lim_{x\to\infty} \left[\frac{x^{100} }{e^{x}}+\left(cos \frac{2}{x}\right)^{x^2}\right] =lim⁡x→∞x100ex+lim⁡x→∞[(cos2x)]x2 = \displaystyle \lim _{x\to \infty } \frac{x^{100} }{e^{x}}+\lim _{x\to \infty }\left[\left(cos \frac{2}{x}\right)\right]^{x^2} =lim⁡x→∞x100ex=0= \displaystyle \lim _{x\to \infty } \frac{x^{100} }{e^{x}} = 0 (Using L' Hopital?? rule) and lim⁡x→∞(cos2x)x2\displaystyle\lim _{x\to \infty }\left(cos \frac{2}{x}\right)^{x^2} is of (1∞)\left(1^{\infty}\right) form =elim⁡x→∞x2(cos2x−1)= e^{\displaystyle \lim_{x \to\infty}} x^{2}\left(cos \frac{2}{x}-1\right) =elim⁡t→0t42(cost−1)= e^{\displaystyle\lim_{t \to 0}} t^{\frac{4}{2}\left(cost-1\right)} (Put 2x=t⇒x=2t\frac{2}{x} = t \Rightarrow x= \frac{2}{t}) =e−lim⁡t→0(1−costt2)4=e−lim⁡t→0(sint2t)4=e−2= e^{\displaystyle -\lim_{t \to 0}} \left(\frac{1-cost}{t^{2}}\right)4 = e^{\displaystyle-\lim_{t \to 0}}\left(\frac{sint}{2t}\right)4 = e^{-2}