Question

Question #f6818

Answer 1

`1/4`

Explanation:

`lim_(x to 0) (2x-sinx)/(3x+sinx)`

indeterminate `0/0`, so applying L'Hopital's rule

`= lim_(x to 0) (2- cos x)/(3+ cosx)`

cos x is continuous through the limit, and the expression overall is continuous, so we can apply x = 0 to the limit

`=(2- 1)/(3+ 1) = 1/4`

Answer 2

Use `lim_(thetararr0)sintheta/theta = 1`

Explanation:

`(2x-sinx)/(3x+sinx)`.
We want to get the expression `sinx/x`. Since the limit at `0` doesn't care what happens when `x = 0` we can multiply by `1` in the form `(1/x)/(1/x)`.

We get

`(2x-sinx)/(3x+sinx) =((1/x)/(1/x)) * ((2x-sinx))/((3x+sinx))`

`= (2-sinx/x)/(3+sinx/x)`

`lim_(xrarr0)(2x-sinx)/(3x+sinx) = lim_(xrarr0)(2-sinx/x)/(3+sinx/x)`

`= (2-(1))/(3+(1))`

`= 1/4`