Question

How do you find `lim sin^2x/x` as `x->0` using l'Hospital's Rule?

Answer 1

`lim_(x->0)(sin^2x)/x=0`

Explanation:

`lim_(x->0)(sin^2x)/x`
If we apply limit then we get `0/0` which is undefined.

Applying L'Hospital Rule
According to this rule we are going to differentiate numerator and denominator w.r.t x and then apply `lim`.

now differentiate numerator and denominator w.r.t x
`lim_(x->0)(2sinxcosx)/1=lim_(x->0)sin2x=sin2(0)=0`

Answer 2

`lim_(x rarr 0) sin^2x/x = 0`

Explanation:

We seek:

`L = lim_(x rarr 0) sin^2x/x`

Both the numerator and the denominator `rarr 0` as `x rarr 0`. thus the limit `L` (if it exists) is of an indeterminate form `0/0`, and consequently, we can apply L'Hôpital's rule to get:

`L = lim_(x rarr 0) (d/dx sin^2x)/(d/dx x)`

`\ \ = lim_(x rarr 0) (2sinxcosx)/(1)`

Which we can now just evaluate to get:

`L = 2sin0cos0`
`\ \ = 2 xx 0 xx 1`
`\ \ = 0`

Note

This particular limit can also be readily evaluated without using L'Hôpital's rule as:

`L = lim_(x rarr 0) sin^2x/x`
`\ \ = lim_(x rarr 0) sinx xx sinx/x`
`\ \ = lim_(x rarr 0) sinx xx lim_(x rarr 0) sinx/x`

The first limit can just be evaluated, and the second limit is a standard calculus result, and is equal to zero. So:

`L = sin 0 xx 0`
`\ \ = 0`