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

##### 1 Answer
Sep 20, 2017

${\lim}_{x \rightarrow 0} \frac{{\int}_{0}^{x} \sin {t}^{2} \mathrm{dt}}{\sin {x}^{2}} = 0$

#### Explanation:

We seek:

$L = {\lim}_{x \rightarrow 0} \frac{{\int}_{0}^{x} \sin {t}^{2} \mathrm{dt}}{\sin {x}^{2}}$

Both the numerator and the2 denominator $\rightarrow 0$ as $x \rightarrow 0$. thus the limit $L$ (if it exists) is of an indeterminate form $\frac{0}{0}$, and consequently, we can apply L'Hôpital's rule to get:

$L = {\lim}_{x \rightarrow 0} \frac{\frac{d}{\mathrm{dx}} {\int}_{0}^{x} \sin \left({t}^{2}\right) \mathrm{dt}}{\frac{d}{\mathrm{dx}} \sin \left({x}^{2}\right)}$

$\setminus \setminus = {\lim}_{x \rightarrow 0} \frac{\frac{d}{\mathrm{dx}} {\int}_{0}^{x} \sin \left({t}^{2}\right) \mathrm{dt}}{\frac{d}{\mathrm{dx}} \sin \left({x}^{2}\right)}$

Now, using the fundamental theorem of calculus:

$\frac{d}{\mathrm{dx}} {\int}_{0}^{x} \sin \left({t}^{2}\right) \mathrm{dt} = \sin \left({x}^{2}\right)$

And,

$\frac{d}{\mathrm{dx}} \sin \left({x}^{2}\right) = 2 x \cos \left({x}^{2}\right)$

And so:

$L = {\lim}_{x \rightarrow 0} \sin \frac{{x}^{2}}{2 x \cos \left({x}^{2}\right)}$

Again this is of an indeterminate form $\frac{0}{0}$, and consequently, we can apply L'Hôpital's rule again to get:

$L = {\lim}_{x \rightarrow 0} \frac{\frac{d}{\mathrm{dx}} \sin \left({x}^{2}\right)}{\frac{d}{\mathrm{dx}} 2 x \cos \left({x}^{2}\right)}$

$\setminus \setminus = {\lim}_{x \rightarrow 0} \frac{2 x \cos \left({x}^{2}\right)}{2 \cos \left({x}^{2}\right) - 4 {x}^{2} \sin \left({x}^{2}\right)}$

Which, we can evaluate:

$L = \frac{0}{2 - 0} = 0$