Question

`lim_(x->0)(tanx-sinx)/(x-sinx)`?

Answer 1

`3`

Explanation:

The limit is of the form `0/0`. We can use L'Hôpital's rule.

Find the derivative of the numerator and denominator separately:
`lim_(x->0)(tan(x)-sin(x))/(x-sin(x))`
`=lim_(x->0)(sec^2(x)-cos(x))/(1-cos(x))`

This is still of the form `0/0`. Use L'Hôpital's rule again:
`=lim_(x->0)((2sin(x))/cos^3(x)+sin(x))/(sin(x))`

Simplify:
`=lim_(x->0)2/cos^3(x)+1`

Just substitute `x=0` to get the answer: `2/cos^3(0)+1=2+1=3`

This is the graph of the function:
graph{(tan(x)-sin(x))/(x-sin(x)) [-16.02, 16.01, -8.01, 8.01]}