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

##### 2 Answers

#### Explanation:

If we apply limit then we get

Applying L'Hospital Rule

According to this rule we are going to differentiate numerator and denominator w.r.t x and then apply

now differentiate numerator and denominator w.r.t x

# 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

# 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#