Question

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

Answer 1

`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`