Find the limit of $\lim_{x \to -\infty}\frac{e^x-\sin(x)}{e^x+\cos^2(x)}$

Question

Check the following limits and if is possible find it:

$$a)\hspace{0.2cm} \lim_{x \to +\infty}\frac{e^x-\sin(x)}{e^x+\cos^2(x)}\hspace{2cm}b)\hspace{0.2cm}\lim_{x \to -\infty}\frac{e^x-\sin(x)}{e^x+\cos^2(x)}$$

$a)\hspace{0.2cm} \lim_{x \to +\infty}\frac{e^x-\sin(x)}{e^x+\cos^2(x)}$

Let $x_n=2\pi n,\, \hspace{0.2cm}x_n\to\infty \ $ and let $y_n=2\pi n+\frac{\pi}{2},\hspace{0.2cm}y_n\to\infty$

$\text{I})\hspace{0.2cm} \lim_{x \to +\infty}\frac{e^x-\sin(x)}{e^x+\cos^2(x)}=\lim_{x_n \to \infty}\frac{e^{2\pi n}-\sin(2\pi n)}{e^{2\pi n}+\cos^2(2\pi n)}=\frac{e^\infty-0}{e^\infty+1}=\frac{\infty}{\infty}\overset{\mathrm{LH}}{=} 1$

$\text{II})\hspace{0.2cm} \lim_{x \to +\infty}\frac{e^x-\sin(x)}{e^x+\cos^2(x)}=\lim_{y_n \to \infty}\frac{e^{2\pi n+\frac{\pi}{2}}-\sin(2\pi n+\frac{\pi}{2})}{e^{2\pi n+\frac{\pi}{2}}+\cos^2(2\pi n+\frac{\pi}{2})}=\frac{e^\infty-1}{e^\infty-0}=\frac{\infty}{\infty}\overset{\mathrm{LH}}{=} 1$

The Limits I and II are equal, then $\hspace{0.2cm}\lim_{x \to +\infty}\frac{e^x-\sin(x)}{e^x+\cos^2(x)}=1$

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For $b)$ I tried to find the limit like $a)$, taking $x_n=-2\pi n, \hspace{0.2cm}y_n=-2\pi n-\frac{\pi}{2}$ but the result was $1$ again, wich is wrong because drawing $\frac{e^x-\sin(x)}{e^x+\cos^2(x)}$ you can see it converges to 1 on the right side and it diverges on the left side.

Is correct my work? How can I find the limit $b$?

Answer 1

I think it's important that not only a correct solution is given, but also that incorrect notions are corrected.

1) The OP seems to be under the impression that substituting a few special series' that converge to $+\infty$ are enough to prove a limit for subproblem a). This is not enough, it has to work for all cases.

2) The right parts of I) and II) (like $\frac{e^\infty - 0}{e^\infty + 1} = \frac\infty\infty = 1)$ seem more of a formally written statement without any actual thought behind them. I also don't know what "LH" stands for (a guess would be L'Hospital)

I come to this conclusion because for problem b) the OP got the right idea, but then claims incorrect limits. It's good that that OP used graphical means to get an idea what to prove. The choice of $x_n=-2\pi n$ and $y_n = -2\pi n -\frac{\pi}2$ is also good.

We have $\lim_{n\to \infty} \frac{e^{x_n} - \sin(x_n)}{e^{x_n} + \cos^2(x_n)} = \lim_{n\to \infty}\frac{e^{-2\pi n} - 0}{e^{-2\pi n} + 1}$

Since $\lim_{n\to \infty}e^{-2\pi n} = 0$, we have to take the limit of a fraction whose enumerator goes to $0$, while the denominator goes to 1. That is easy, we just apply the limit quotient rule and get $\lim_{n\to \infty} \frac{e^{x_n} - \sin(x_n)}{e^{x_n} + \cos^2(x_n)} = 0$.

Similiarly, we get $\lim_{n\to \infty} \frac{e^{y_n} - \sin(y_n)}{e^{y_n} + \cos^2(y_n)} = \lim_{n\to \infty}\frac{e^{-2\pi n - \frac{\pi}2} + 1}{e^{-2\pi n - \frac{\pi}2} + 0}$.

In this case the enumerator goes to 1, while the denominator goes to 0. By the limit quotient rule, that means the limit diverges to $\infty$ (this analysis is not enough to find out if the quotient gets both arbitrarily high and arbitrarily low; but it is clear that at least the absolute value gets arbitrarily high).

So we have found one special series where the limit in b) is 0 and another where it is $\infty$. So, as the OP had already realized, the limit in b) does not exist (not even in the sense that if may be $\infty$).

Answer 2

Consider doing this.

$$\lim_{x \to +\infty}\frac{e^x-\sin(x)}{e^x+\cos^2(x)} =\lim_{x \to +\infty}\frac{1-e^{-x}\sin(x)}{1+e^{-x}\cos^2(x)} $$ Now what happens as $x\to\infty$?

Answer 3

For the limit at $\infty$, rewrite it as $$ \lim_{x\to\infty}\frac{e^x\left(1-\dfrac{\sin x}{e^x}\right)} {e^x\left(1+\dfrac{\cos^2x}{e^x}\right)} $$ For the limit at $-\infty$, it's easier if you do $x=-t$ and consider $$ \lim_{t\to\infty}\frac{e^{-t}+\sin t}{e^{-t}+\cos^2t} $$ Now try to see what happens with $t=n\pi$ and $t=(2n+1)\pi/2$.