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