$\lim_{x\to \infty}A(x)$

Question

If $A(x)=\int_{-1}^{x}e^{-|t|}dt$, then what is $\lim_{x\to \infty}A(x)$?

I am not able to make any progress on this problem.I hope someone can help me with this.

Answer 1

$\int_{-1}^{x}e^{-|t|}dt$=$\int_{-1}^{0}e^{t}dt$+$\int_{0}^{x}e^{-t}dt$. This is because |t|=-t if t<0 and |t|=t if $t{>=}0$.
$\int_{-1}^{0}e^{t}dt$=1-1/e and $\int_{0}^{x}e^{-t}dt$=-$\int_{0}^{-x}e^{t}dt$=-($e^{-x}$-1)

Thus $\int_{-1}^{x}e^{-|t|}dt$=$=2-1/e-$$e^{-x}$

In the limit x->inf

A=2-1/e

Answer 2

If $x\ge 0$ then $$ A(x)=\int_{-1}^x e^{-t}\,dt = \int_{-1}^0 e^t\,dt + \int_0^x e^{-t}\,dt. $$ Integrate, then find $\lim\limits_{x\to\infty}$.