Question

Question #6c5f5

Answer 1

`Lim_(xrarr0^+)(e^(3x)-5x)^(1/x)=e^-2`

Explanation:

`Lim_(xrarr0^+)(e^(3x)-5x)^(1/x)`

We have to apply Euler's identity: `e^lnx=x`
Then we get this:

`e^(Lim_(xrarr0^+)ln(e^(3x)-5x)^(1/x))`

To make it look simple let's evaluate limit first

`Lim_(xrarr0^+)ln(e^(3x)-5x)^(1/x)=Lim_(xrarr0^+)1/xln(e^(3x)-5x)`

`Lim_(xrarr0^+)(ln(e^(3x)-5x))/x=0/0`

Using L'Hopitals rule:

`Lim_(xrarr0^+)(1/(e^(3x)-5x)*(3e^(3x)-5))/1`

`Lim_(xrarr0^+)(3e^(3x)-5)/(e^(3x)-5x)=(3*1-5)/(1-5*0)=-2/1=-2`

So the answer is: `e^-2`