The limit $\lim_{x \to ∞} (3x^4-5x^2+1)(1-e^{1/x^2})^2$ $\lim_{x \to ∞} (3x^4-5x^2+1)(1-e^{1/x^2})^2$
I found +∞ but the limit is 3. How to compute this limit ?

$\displaystyle \lim_{ x \to \infty} (3x^4-5x^2+1)(1-e^{1/x^2})^2$
$=\displaystyle \lim_{x \to \infty} \left(\frac{e^{1/x^2}-1}{\frac{1}{x^2}}\right)^2\left(\frac{3x^4-5x^2+1}{x^4}\right)$
$=1\times 3$
You know that: $$\lim_{u \to 0^+} \frac{e^u-1}{u}=1$$
Putting $u=1/x^2$ we get:$$\lim_{x \to \infty} \frac{e^{\frac{1}{x^2}}-1}{\frac{1}{x^2}}=1$$
