Question

What is `lim x->0` of `(sin4x)/(sin5x)`?

Answer 1

`4/5`

Explanation:

`(sin4x)/(sin5x)= 4/5(5x)/(4x)(sin4x)/(sin5x)=4/5((sin4x)/(4x))/((sin5x)/(5x))`

now

`lim_(x->0)(sin4x)/(sin5x)=lim_(x->0)4/5((sin4x)/(4x))/((sin5x)/(5x))=4/5`

Answer 2

`lim_(xrarr0)(sin4x)/(sin5x) = 4/5`

Explanation:

The limit is of the indeterminate form `0/0` so we can use L'Hôpital's rule as follows;

`lim_(xrarr0)(sin4x)/(sin5x) = lim_(xrarr0)(d/dxsin4x)/(d/dxsin5x)`
`:. lim_(xrarr0)(sin4x)/(sin5x) = lim_(xrarr0)(4cos4x)/(5cos5x)`
`:. lim_(xrarr0)(sin4x)/(sin5x) = 4/5lim_(xrarr0)(cos4x)/(cos5x)`

Then, as both `cos4xrarr1` and `cos5xrarr1` as `xrarr0` we have;

`lim_(xrarr0)(sin4x)/(sin5x) = 4/5`,