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

2 Answers
Sep 22, 2017

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

Sep 22, 2017

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