How to solve $\lim\limits_{x \to -\infty } e^{x^2/(1+x)}$?

Question

so I tried to solve this limit as it follows:

$$\lim\limits_{x \to -\infty } e^{x^2/x(1/x + 1)} = \lim\limits_{x \to -\infty } e^{x/(1/x + 1)} = \infty $$

.. But the solution is 0 Can someone help me please? Thank you in advance!

Answer 1

$\lim\limits_{x \to -\infty } {{x^2}\over{1+x}}=-\infty$ implies the limit is zero.

Answer 2

Note that for $x\to -\infty$

$$e^{\frac{x^2}{1+x}} \to e^{-\infty}=0$$

indeed

$$\frac{x^2}{1+x}=x \cdot\frac{x}{1+x}\to-\infty\cdot1=-\infty$$

Answer 3

Since the exponential map is continous, it commutes with the limit, i.e.

$$ \lim_{x\to a} e^{f(x)} = e^{\lim\limits_{x\to a}f(x)}. $$

In your case, $f(x)=\frac{x^2}{1+x}$ and hence:

$$ \lim_{x\to -\infty} e^{\frac{x^2}{1+x}} = e^{\lim\limits_{x\to -\infty}\frac{x^2}{1+x}} = e^{-\infty} = 0. $$