Question

Question #c91de

Answer 1

1

Explanation:

assuming you mean natural log here

`lim_{x to 0} (ln (1+x)) / x`

`=lim_{x to 0} (x - x^2/2 + mathcal(O)(x^3) ) / x` using the Taylor expansion for the numerator about x = 0

`=lim_{x to 0} 1 + mathcal(O)(x) = 1`

of you can hit it with L'Hopital as it's `0/0` indeterminate so

`lim_{x to 0} ((1/(1+x))) / 1 = 1/(1+x) = 1`

you can also look at it as

`lim_{x to 0} 1/xln (1+x)`

`lim_{x to 0} ln color(red)((1+x)^(1/x))`

and that bit in red is the definition of `e` - see Bernoulli's compound interest formula, well-covered on Wiki, so you have `ln e = 1`

you'd need a little bit more algebra to do that, ie lift the `ln` outside the limit, which you can do as it is continuous about the limit

Answer 2

`1`

Explanation:

We need to find
`lim_(x->0) log (1+x)/x`
Substituting `x=0` in the expression `log (1+x)/x`, we see that it is indeterminate as in the form `0/0 .`
Therefore apply L'Hospital's Rule. Differentiate the numerator and differentiate the denominator and then take the limit.

We obtain
`lim_(x->0) (1/ (1+x))/1`
`=>lim_(x->0) 1/ (1+x)`
`= 1/ (1+0)`
`= 1`