Question

How to evaluate limits `[e^x-(1-x)]/x` as x approaches 0?

`[e^x-(1-x)]/x`

Answer 1

`lim_(x->0)(e^x-(1-x))/x=2`

Explanation:

As `x->0`, `(e^x-(1-x))/x` appears as `0/0`

hence, to find its limits, we can use L'Hospital's rule

according to which

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

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

= `2/1=2`

Answer 2

Recall that `e^x = 1 + x + x^2/2 + x^3/6 + ...`:

`L = lim_(x-> 0) (1 + x + x^2/2 + x^3/6 + ... -1 + x)/x`

`L = lim_(x-> 0) (2x + x^2/2 + x^3/6 + ...)/x`

`L = lim_(x-> 0) 2 + x/2 + x^2/6 + ...`

`L = 2`

The graph confirms:

Hopefully this helps!