`f(x)={(e^x-1)/(log(1+2x)),\ \ \ if\ x!=0\ \ \ \ \ \ \ \ \ 7,\ \ \ \ \ \ \ \ \ \ \ if\ x=0` at `x=0`

Explanation

We have,

`f(x)= frac{e^{x}-1}{log (1+2 x)}, ` if ` x equiv 0 `
7, if x=0
We see that,
`
lim _{x rightarrow 0} f(x)`


`=lim _{x rightarrow 0} frac{e^{x}-1}{log (1+2 x)} `

`=lim _{x rightarrow 0} frac{e^{x}-1}{(frac{2 x log (1+2 x)}{2 x})}`
`
...