What is the value of? #lim_(x->0)(int_0^x sin t^2.dt)/sin x^2#

1 Answer
Steve M
Sep 20, 2017

# lim_(x rarr 0) (int_0^x sin t^2 dt)/(sin x^2) = 0#

Explanation:

We seek:

# L = lim_(x rarr 0) (int_0^x sin t^2 dt)/(sin x^2)#

Both the numerator and the2 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 int_0^x sin (t^2) dt)/(d/dx sin (x^2) ) #

# \ \ = lim_(x rarr 0) (d/dx int_0^x sin (t^2) dt)/(d/dx sin (x^2) ) #

Now, using the fundamental theorem of calculus:

# d/dx int_0^x sin (t^2) dt = sin(x^2) #

And,

# d/dx sin(x^2) = 2xcos(x^2)#

And so:

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

Again this is of an indeterminate form #0/0#, and consequently, we can apply L'Hôpital's rule again to get:

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

# \ \ = lim_(x rarr 0) (2xcos(x^2))/(2cos(x^2)-4x^2sin(x^2) ) #

Which, we can evaluate:

# L = (0)/(2-0 ) = 0#

Related questions
Impact of this question
10644 views around the world
You can reuse this answer
Creative Commons License