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How would one find the limit for the following problem.

$x\rightarrow\infty$

$\frac{e^{x}+x-\cos(2x)}{x^2}$

I did the hospital rule.

$\frac{e^x+1+2\sin(2x)}{2x}$

But now I am stuck I did this but I feel it diverges.

$e^x+1+2\sin(2x)*\frac{1}{2x}$

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  • $\begingroup$ As a general observation you should suspect it diverges because at $+\infty$ the numerator is ruled by $e^x$ which crushes $x^2$. $\endgroup$ – Git Gud Oct 30 '13 at 17:30
  • $\begingroup$ Are you sure that $x$ tends towards $\infty$ and not towards $0$ ? $\endgroup$ – Lucian Oct 30 '13 at 17:44
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$$\lim_{x\to\infty}\frac{e^x+x-\cos 2x}{x^2}\stackrel{\text{l'H}}=\lim_{x\to\infty}\frac{e^x+1+2\sin 2x}{2x}\stackrel{\text{l'H}}=\lim_{x\to\infty}\frac{e^x+4\cos 2x}2=\infty$$

The limit thus exists (in the wide sense of the word), since $\;\cos 2x\;$ is bounded.

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  • $\begingroup$ hmm this seems to make sense because cos(2x) is always between -1<x<1 like this. $\endgroup$ – Fernando Martinez Oct 30 '13 at 17:38
  • $\begingroup$ "Seems to make sense"? Bueno...:) $\endgroup$ – DonAntonio Oct 30 '13 at 17:40
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Hint: $$e^x > \dfrac{x^3}{3!}, \qquad x \geqslant 0.$$

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Since $\cos(2x) \leq 1,\,\forall x$, and $e^x = 1+x+x^2/2+x^3/6+\cdots \geq 1+x+x^2/2+x^3/6$, we obtain, $$\frac{e^{x}+x-\cos(2x)}{x^2} \geq \frac{2x+x^2/2+x^3/6}{x^2} \geq \frac{x^3/6}{x^2} \geq \frac{x}{6},\,\forall x > 0.$$ The lower bound diverges, hence the original series.

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Solution 1: Apply d'Hospital rule once more to the result.

Solution 2: $e^x$ grows faster than any power of $x$. Thus, you immediately see that the original sequence diverges.

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