Question

How do you evaluate the limit `(3e^x-sinx+ln(x+1))` as x approaches `0`?

Answer 1

I got `3`

Explanation:

I would try by substituting `x=0` into our function `f(x)` getting:

`f(0)=3e^0-sin(0)+ln(0+1)=3-0-0=3`

so that:

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

Graphically we can observe the behavior of our function around `x=0`: