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I would like to compute the limit of:

$$\lim_{x\to -\infty} f(x)$$

where

$$ {f(x)=e^{-2x}(\cos(x)-\sin(x))}$$

My thought process was that since cosx and sinx are bounded between 1 and -1 and the exponential function goes to infinity the function will not converge. Is that correct and how can I prove it?

Thanks!

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Yes, it will diverge. –  Peter Horvath Mar 14 at 13:36
    
Yes that is correct! To formally prove it show that for any $\epsilon>0$ you can find an $N\in\mathbb{R}$ such that $|f(x)|>\epsilon$ for all $x<N$. –  Marc Mar 14 at 13:41
    
Hint : look at the sign of $\cos(x)-\sin(x)$ and the graph of $\tan(x) - 1$. You may see why there is a problem with a potential limit of $+\infty$ or $-\infty$ –  T_O Mar 14 at 13:49

4 Answers 4

up vote 0 down vote accepted

If diverge is the logical negation of converge, then you are right. But if diverge means divergent to infinity, then I am sorry but you are wrong. The limit does not exist at all, as WoframAlpha clearly shows.

For example, can you find a sequence $x_n \to -\infty$ such that $\cos x_n=\sin x_n$? Along this sequence, $f(x_n)=0$. But you can easily find sequences $\{x_n\}$ such that $f(x_n) \to \pm\infty$.

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It is divergent: $$ f(x)=\sqrt{2}e^{-2x}\cos(x+\frac{\pi}{4}) $$

The envelope of this function goes to infinity, but the function itself oscillates between minus infinity and plus infinity.

Simply put: no limit!

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Yes, it is infinite.

put $y = -x$

$ \lim_{x \to -\infty} \implies \lim_{y \to \infty}$.

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Yes, it diverges. Show that, for any $M<0$, there is an $N_1<M$ with $f(N_1)<-1$ and an $N_2<M$ with $f(N_2)>1$. That means it can't settle down to a single value.

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