Question

What is `lim_(x->0) ln(sin x + cos x)/x` ?

Answer 1

`lim_(x->0) ln(sin x + cos x)/x = 1`

Explanation:

`color(white)()`
Intuitive analysis

Notice that for small values of `x`:

`sin x ~~ x`

`cos x ~~ 1`

`ln (x+1) ~~ x`

So:

`ln(sin x+cos x) ~~ ln(x+1) ~~ x`

So we should expect:

`lim_(x->0) ln(sin x + cos x)/x = 1`

`color(white)()`
Semi-formal analysis

`sin x = sum_(n=0)^oo ((-1)^n)/((2n+1)!) x^(2n+1) = x+O(x^3)`

`cos x = sum_(n=0)^oo ((-1)^n)/((2n)!) x^(2n) = 1+O(x^2)`

`ln (x + 1) = sum_(n=0)^oo ((-1)^n)/(n+1) x^(n+1) = x+O(x^2)`

So:

`ln(sin x + cos x)/x = ln(x+1+O(x^2))/x`

`color(white)(ln(sin x + cos x)/x) = (x+O(x^2))/x`

`color(white)(ln(sin x + cos x)/x) = 1+O(x) -> 1" "` as `x -> 0`