Limit of e^(-2x) * cosx as x-> ∞

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Problem statement: Find the limit of e^(-2x) * cosx as x-> ∞

Attempt at solution: I tried to say that the limit of e^(-2x) as x-> ∞ is 0, and the limit of cosx doesn't exist since it oscillates between -1 and 1 therefore the overall answer is that it doesn't exist. I was wrong however and the right answer was 0. Could someone please help?
 
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limit of cos(x) doesn't exist but can you put bounds on its absolute value?

whats the limit of e^(-2x) times the upper bound?
 
Try applying the squeeze theorem.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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