Question

How do you find `\lim _ { x \rightarrow 1} \frac { e ^ { 2x } + e ^ { x } } { e ^ { x } - 1}`?

Answer 1

`lim_(x->1)(e^(2x)+e^x)/(e^x-1)=(e^2+e)/(e-1)~~5.88224`

Explanation:

Use direct substitution. Plug in `1` for every `x` value:

`lim_(x->1)(e^(2x)+e^x)/(e^x-1)=(e^(2(1))+e^(1))/(e^(1)-1)=(e^2+e)/(e-1)~~5.88224`