# Finding the limit of $\frac{e^{x}+x-\cos(2x)}{x^2}$

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.

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

• As a general observation you should suspect it diverges because at $+\infty$ the numerator is ruled by $e^x$ which crushes $x^2$. – Git Gud Oct 30 '13 at 17:30
• Are you sure that $x$ tends towards $\infty$ and not towards $0$ ? – Lucian Oct 30 '13 at 17:44

$$\lim_{x\to\infty}\frac{e^x+x-\cos 2x}{x^2}\stackrel{\text{l'H}}=\lim_{x\to\infty}\frac{e^x+1+2\sin 2x}{2x}\stackrel{\text{l'H}}=\lim_{x\to\infty}\frac{e^x+4\cos 2x}2=\infty$$

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

• hmm this seems to make sense because cos(2x) is always between -1<x<1 like this. – Fernando Martinez Oct 30 '13 at 17:38
• "Seems to make sense"? Bueno...:) – DonAntonio Oct 30 '13 at 17:40

Hint: $$e^x > \dfrac{x^3}{3!}, \qquad x \geqslant 0.$$

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.

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.