Question

How do you find the limit of `(e^x + sin(x) + x - 1)/(e^x - 1)` as x approaches 0?

Answer 1

`lim_{x->0}(e^x+sin(x)+x-1)(e^x-1) = 3`

Explanation:

`f(x)=(e^x+sin(x)+x-1)(e^x-1) = (e^x-1)/(e^x-1)+(sin(x)+x)/(e^e-1)`

so

`f(x)=1+(x(sin(x)/x+1))/(x(e^x-1)/x) = 1 + (sin(x)/x+1)/((e^x-1)/x)`

Here `lim_{x->0}sin(x)/x=1` and `lim_{x->0}(e^x-1)/x=1`

Putting all together

`lim_{x->0}f(x)=1+2/1=3`