Question

`lim_(xrarr0)(e^(2x)+x)^(1//x)=` ?

Answer 1

`lim_(xrarr0)(e^(2x)+x)^(1//x)=e^3`

Explanation:

`L=lim_(xrarr0)(e^(2x)+x)^(1//x)`

To undo powers like this in limits, take the natural logarithm of both sides.

`ln(L)=ln(lim_(xrarr0)(e^(2x)+x)^(1//x))`

The logarithm, being a continuous function, can be moved inside the limit.

`ln(L)=lim_(xrarr0)ln((e^(2x)+x)^(1//x))`

Move the power outside of the logarithm using log rules.

`ln(L)=lim_(xrarr0)ln(e^(2x)+x)/x`

Notice that as `xrarr0`, we have this limit in the indeterminate form `0/0`. That means L'Hospital's rule applies, and we can take the derivatives of the numerator and denominator separately in the limit.

`ln(L)=lim_(xrarr0)(d/dx(ln(e^(2x)+x)))/(d/dx(x))`

`ln(L)=lim_(xrarr0)((2e^(2x)+1)/(e^(2x)+x))/1`

`ln(L)=lim_(xrarr0)(2e^(2x)+1)/(e^(2x)+x)`

We can now evaluate the limit.

`ln(L)=(2e^0+1)/(e^0+0)`

`ln(L)=(2+1)/1`

`ln(L)=3`

`L=e^3`

Answer 2

`e^3`

Explanation:

`(e^(2x)+x)^(1/x)= e^2(1+x/e^(2x))^(1/x)` now, making `y = x/e^(2x)`
`(e^(2x)+x)^(1/x)=e^2(1+y)^(1/(y e^(2x))) = e^2((1+y)^(1/y))^(1/(e^(2x)))`

Now `lim_(x->0) =>lim_(y(x)->0)` so

`lim_(x->0)(e^(2x)+x)^(1/x)=e^2lim_(x->0)(lim_(y->0)(1+y)^(1/y))^(1/e^(2x))=e^2 xx e=e^3`