We have $f:RR->RR,f(x)=(x+2)e^(-|x|)$ How to solve this limit? $lim_(x->0)(f(x)-f(0))/(x);x>0$

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Answer 1 · 2017-04-17T17:19:49

The limit does not exist.

The left and right limits at $0$ are different.

The quickest way to see this is to note that this limit is the definition of $f'(0)$ .

But for this function we have

$f(x) = {((x+2)e^x,"if",x >= 0),((x+2)e^(-x),"if",x < 0):}$

The right derivative is $e^x+(x+2)e^x$ which is $2$ at $x=0$

while the left derivative is $e^(-x)-(x+2)e^-x$ , which is $-2$ and $x=0$ .

Using technology

The graph of $f$ is shown below.

graph{y=(x+2)e^(-abs(x)) [-5.213, 5.884, -2.11, 3.44]}

We have f:RR->RR,f(x)=(x+2)e^(-|x|)f:RR->RR,f(x)=(x+2)e^(-|x|)How to solve this limit? lim_(x->0)(f(x)-f(0))/(x);x>0lim_(x->0)(f(x)-f(0))/(x);x>0