# Find the limit of the function

Function: $e^{-2x}x\cos(x^2)$ if x tends to +infinity. I've tried to calculate it but I got 0*infinity, it's indeterminate.

• Do you mean $$\lim_{x \to \infty}e^{-2x}xcos(x^2)=?$$ – Khosrotash Sep 27 '16 at 18:33
• hint $\cos(x)\leq 1$ – tired Sep 27 '16 at 18:36

## 2 Answers

Since $\cos(x^2)$ is bounded and $\lim_{x\to\infty}xe^{-2x}=0$:

$$\lim_{x\to\infty}x\cos(x^2)e^{-2x}=0$$

Since $-1\leq\cos(x^2)\leq 1$ for all $x$ we have $$0 = \lim_{x\rightarrow\infty}-xe^{-2x}\leq \lim_{x\rightarrow\infty}\cos(x^2)x e^{-2x}\leq \lim_{x\rightarrow\infty}xe^{-2x} = 0$$ so $\lim_{x\rightarrow\infty}\cos(x^2)x e^{-2x} = 0$ by the squeeze theorem