Complex limit question

Question

Calculate $\displaystyle\lim_{x\to\infty} \frac{x^2e^{-4x}+xe^{-3x}}{e^{-2x}+xe^{-3x}}$.

Calculate $\displaystyle\lim_{x\to 0} \frac{x^2e^{x^4}-\sin{(x^2)}}{1-\cos{(x^3)}}$.

For the first question I got \begin{equation*} \begin{split} \lim_{x\to\infty} \frac{x^2e^{-4x}+xe^{-3x}}{e^{-2x}+xe^{-3x}} &= \lim_{x\to\infty} \frac{xe^{-3x}\left(xe^{-x}+1\right)}{e^{-2x}\left(xe^{-x}+1\right)} \\ &= \lim_{x\to\infty} \frac{xe^{-3x}}{e^{-2x}} \\ &= \lim_{x\to\infty} \frac{x}{e^x}. \end{split} \end{equation*} This has an indeterminate form $\frac{\infty}{\infty}$. So we can use L'Hopital's rule to get $\lim_{x\to\infty} \frac{x}{e^x} = \lim_{x\to\infty} \frac{1}{e^x} = 0$.

For the second question I tried L'Hopital's rule but that didn't work and I also tried multiplying by the conjugate $\left(1+\cos{(x^3)}\right)$. Any help would be great!!!

For the 2nd Q use $1-\cos A=2(\sin A/2)^2 ,$ which is obtained from $\cos A= 2(\cos A/2)^2-1=2(1-(\sin A/2)^2)-1.$ — DanielWainfleet, Jul 30 '19 at 10:22

José Carlos Santos · Accepted Answer · 2019-07-29T22:27:22.343

1

Near $0$, you have\begin{align}x^2e^{x^4}-\sin(x^2)&=x^2\left(1+x^4+\frac{x^8}{2!}+\cdots\right)-\left(x^2-\frac{x^6}6+\cdots\right)\\&=\frac76x^6+O(x^7)\end{align}and$$1-\cos(x^3)=\frac{x^6}2+O(x^7)$$and therefore$$\lim_{x\to0}\frac{x^2e^{x^4}-\sin(x^2)}{1-\cos(x^3)}=\frac{7/6}{1/2}=\frac73.$$

edited Jul 29 '19 at 22:27

answered Jul 29 '19 at 22:15

José Carlos Santos

427,504

Thank you very much I didn't think to use a series expansion!!! – squenshl Jul 29 '19 at 22:30
@José Carlos Santos, how were you able to obtain $O(x^7)$. That has got me thinking for quite a number of times. – Nzewi Ernest Kenechukwu Aug 06 '19 at 08:22
If $\sum_{n=0}^\infty a_nx^n$ is a power series and $N\in\mathbb N$, then$$\sum_{n=0}^\infty a_nx^n=a_0+a_1x+\cdots+a_{N-1}x^{N-1}+O(x^N).$$ – José Carlos Santos Aug 06 '19 at 08:33

Complex limit question

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