Somewhat complicated limit (no l'Hopital permitted)

Question

I'm trying to calculate the following limit:

$$ \lim_{x\to +\infty}\frac{f(x)}{g(x)} $$ where $$ f(x) = e^{-2x}(\cos(x)+2\sin(x))$$ and $$g(x) = e^{-x}(\cos(x)+\sin(x))$$

Now, because $g'$ has zero points, we can't use the de l'Hopital rule, but when I draw the graph, it does seem that de l'Hopital would produce a valid answer ($0$).

How would I go about calculating this limit? (My intuition says it doesn't exist.)

Answer 1

L'Hôpital's rule is not normally used when finding a limit as an expression tends to $\infty$.

In your case, $$ g(x)=e^{-x}(\cos(x)+\sin(x))=\sqrt{2}e^{-x}\sin\left(x+\frac\pi4\right) $$ so $g$ has zeroes at $-\frac\pi4,\frac{3\pi}4,\frac{5\pi}4,\dots$. Therefore, the expression $$ \frac{f(x)}{g(x)} $$ is not defined at these points. Indeed, since $f$ does not have a zero at $-\frac\pi4,\frac{3\pi}4,\frac{5\pi}4,\dots$, there is no way we could make $f/g$ into a continuous function even if we wanted to. So there is no limit as $x\to+\infty$.

Answer 2

Note that if $\tan x = -\dfrac{1}{2}$ then $f(x) = 0$, and if $\tan x = -1$ then $g(x) = 0$. Thus the ratio $\dfrac fg$ alternates between $0$ and undefined infinitely often as $x \to \infty$ and can't have a limit.

Answer 3

$$\lim_{x \to \infty} \frac{e^{-2x}(\cos x + 2 \sin x)}{e^{-x}(\cos x + \sin x)}= \lim_{x \to \infty}\frac{\sqrt 5(\cos x - \tan^{-1}(2))}{\sqrt 2e^{x}\cos (x + \pi/4)} = \text{does not exist}$$

Answer 4

The limit does not exist because of the periodic root in the denominator (that does not coincide with a root in the numerator): it is such that for $x$ as large as you want, it will never be possible to bracket the function.