Solving the Limit of (ex+sinx)/sinx as x->0

[Dell]

how do i solve this limit

Lim (e^x+sinx)^1/sinx
x->0

its almost oiler (1+x)^1/x, at 1st looking at it i thought it was simple e, because e^x when x->0 is 1, but i see i need to somehow turn the e^x into 1,,, any ideas??

 

[NoMoreExams]

I would take ln of both sides and then use L'Hopital's

 

[Dell]

both of which sides?? can you show/describe what you mean

 

[tim_lou]

take the ln and use L'Hopital.

If you just want to evaluate it, expand it first order in x should be good.

Be careful though, for small x,
e^x\approx 1+x
and
\sin(x) \approx x
so your "intuitive" conclusion (1+x)^{1/x} doesn't hold.

 

[Dell]

what do i do with the ln,?? take ln on what??

 

[NoMoreExams]

You have

y = \lim_{x \rightarrow 0} \left(e^{x} + sin(x)\right)^{\frac{1}{sin(x)}}

If you take ln of both sides of the expression you have

ln(y) = \lim_{x \rightarrow 0} \frac{ln(e^{x} + sin(x))}{sin(x)}

Now do L'Hopital's and go from there, remember in the end you want to solve for y.