How to compute this type of limit? Good evening to everyone. The limit that I have to compute is the fallowing: $ \lim\limits _{x\to -1}\left(1-\left|x\right|\right)e^{\frac{1}{2x+2}} $ . The right answers are: $$ \lim _{x\to -1^+}\left(1-\left|x\right|\right)e^{\frac{1}{2x+2}} = \infty $$ and $$ \lim _{x\to -1^-}\left(1-\left|x\right|\right)e^{\frac{1}{2x+2}} = 0$$. 
Edit: I tried limit substitution but it doesn't work. Nor using the $e^{\ln(x)}$ strategy.
Thanks for any possible responses.
 A: When you do a limit substitution, you also need to change the value the variable approaches. As $x \to -1$, since $y=2x+2$, $y \to 2(-1)+2=0$. Therefore, we get the limit:
$$\lim_{y \to 0}\left(1-\left\lvert \frac{y-2}{2} \right\rvert\right)e^{\frac 1 y}$$
Now, we can not just substitute $y=0$ into the function because then we get $e^{\frac 1 0}$, which is undefined.
However, once we have this, it's easier to analyze because the exponent is simpler. First, notice that:
$$\lim_{y \to 0}1-\left\lvert \frac{y-2}{2} \right\rvert=0$$
This is true for both sides.

For the negative side, we need to look at
$$\lim_{y \to 0^-}e^{\frac 1 y}$$
As $y \to 0^-$, $\frac 1 y \to -\infty$ which means that this is really:
$$\lim_{z \to -\infty} e^z=0$$
Thus, for the negative side, we have:
$$\lim_{y \to 0^-}\left(1-\left\lvert \frac{y-2}{2} \right\rvert\right)e^{\frac 1 y}=\left(\lim_{y \to 0^-}1-\left\lvert \frac{y-2}{2} \right\rvert\right)\left(\lim_{y \to 0^-}e^{\frac 1 y}\right)=0\cdot 0=0$$

Now, for the positive side, we need to look at
$$\lim_{y \to 0^+}e^{\frac 1 y}$$
As $y \to 0^+$, $\frac 1 y \to \infty$ which means that this is really:
$$\lim_{z \to \infty} e^z=\infty$$
Thus, for the positive side, we have:
$$\lim_{y \to 0^+}\left(1-\left\lvert \frac{y-2}{2} \right\rvert\right)e^{\frac 1 y}=\left(\lim_{y \to 0^+}1-\left\lvert \frac{y-2}{2} \right\rvert\right)\left(\lim_{y \to 0^+}e^{\frac 1 y}\right)=0\cdot \infty$$
This is an indeterminate form. Usually, when I see things like this, I immediately go to L'Hopital's rule. We can do this by turning the limit into a fraction as so:
$$\lim_{y \to 0^+}\left(1-\left\lvert \frac{y-2}{2} \right\rvert\right)e^{\frac 1 y}=\lim_{y \to 0^+}\frac{\left(1-\left\lvert \frac{y-2}{2} \right\rvert\right)}{e^{-\frac 1 y}}$$
Now, the numerator still goes to $0$, but the denominator now has a negative exponent, so it becomes $\lim_{z \to -\infty} e^z=0$, so the limit is now in the form of $\frac 0 0$. Therefore, we can apply L'Hopital's rule.
Now, what is the derivative of:
$$1-\left\lvert \frac{y-2}{2} \right\rvert$$
If you graph this function and look at $y=0$, it becomes clear that the slope of the line is $\frac 1 2$.
Meanwhile, the derivative of $e^{-\frac 1 y}$ can be found using chain rule:
$$\frac{d}{dy}\left(e^{-\frac 1 y}\right)=\frac{d}{dy}\left(-\frac 1 y\right)\frac{d}{d\left(-\frac 1 y\right)}\left(e^{-\frac 1 y}\right)=\frac{1}{y^2}e^{-\frac 1 y}$$
Thus, by L'Hopital's rule, we get:
$$\lim_{y \to 0^+}\frac{\left(1-\left\lvert \frac{y-2}{2} \right\rvert\right)}{e^{-\frac 1 y}}=\lim_{y \to 0^+}\frac{\frac 1 2}{\frac{1}{y^2}e^{-\frac 1 y}}=\frac 1 2\lim_{y \to 0^+}\frac{y^2}{e^{-\frac 1 y}}=\frac 1 2\lim_{y \to 0^+}y^2e^{\frac 1 y}$$
Now, we break the limit up into $y^2$ and $e^{\frac 1 y}$:
$$\frac 1 2\left(\lim_{y \to 0^+}y^2\right)\left(\lim_{y \to 0^+}e^{\frac 1 y}\right)=\frac 1 2\infty\cdot \infty=\infty$$
A: I hope it is clear why
$\lim_\limits{x\to-1^-} (1-|x|) e^{\frac{1}{2x+2}} = 0$
As x approaches $-1$ from the left ${\frac{1}{2x+2}}$ is going to $-\infty$ and $(1-1)e^{-\infty} = (0)(0) = 0$
$\lim_\limits{x\to-1^+} (1-|x|) e^{\frac{1}{2x+2}}$
In a neighborhood of $-1, |x| = -x,$ and let $y = \frac {1}{x+1}$
And, use L'Hopitals rule or a Taylor series expansion to show that 
$\lim_\limits{y\to \infty} \dfrac {e^{\frac y2}}{y} = \infty$
A: (1). For $-1< x<0$ let $x=-1+y$ with $0<y<1.$ Then $1-|x|=y$ and $e^{1/(2 x+2)}=e^{1/2 y}.$ 
Let $y=1/2 z.$ Then $(1-|x|)e^{1/(2 x+2}=e^z/2 z.$ 
Now $z\to \infty$ as $x\to -1^+$ so it suffices to know that $\lim_{z\to \infty}e^z/z=\infty.$ 
(2). The case $x<-1$ is handled similarly. Let $x=-1-v$ with $v>0$ and let $v=1/2 u.$ (Etc.) It suffices to know that $\lim_{u\to \infty}e^{-u}=0.$
