Question

If `L = lim_(x-> 0) (e^x - 1)/x`, what is the value of `L`?

Answer 1

Use a Mclaurin series

Explanation:

The taylor series for `e^x` is `1 + x + x^2/(2!) + x^3/(3!) + ...` at `x = 0`. Hence, we can substitute this into the limit.

`L = lim_(x->0) (1 + x + x^2/(2!) + x^3/(3!) + ... - 1)/x`

`L = lim_(x->0) (x + x^2/(2!) + x^3/(3!) + ...)/x`

`L = lim_(x->0) x/x + x^2/(x2!) + x^3/(x3!) + ...`

`L = lim_(x->0) 1 + x/(2!) + x^2/(3!) + ...`

`L = 1 + 0 + 0 + 0 + ...`

`L = 1`

So the limit has been proven.

Hopefully this helps!

Answer 2

See below.

Explanation:

Making `h = x` we have

`lim_(h->0)(e^h-1)/h = lim_(h->0)(e^(0+h)-e^0)/h = (d/(dx)e^x)(0) = e^0 = 1`