Exercise:
Find $\underset{x\to 1}{\mathop{\lim }}\,\frac{\int_{1}^{x}{\sin
\left( {{t}^{2}}-1 \right)dt}}{\sin \left( \int_{1}^{x}{\left( {{t}^{2}}-1
\right)dt} \right)}$
Solution: Let $f\left( x \right)=\int_{1}^{x}{\sin \left( {{t}^{2}}-1
\right)dt\,\,\And \,\,g\left( x \right)=\int_{1}^{x}{\left( {{t}^{2}}-1
\right)dt}}$
So $\underset{x\to 1}{\mathop{\lim }}\,\frac{\int_{1}^{x}{\sin \left(
{{t}^{2}}-1 \right)dt}}{\sin \left( \int_{1}^{x}{\left( {{t}^{2}}-1 \right)dt}
\right)}=\underset{x\to 1}{\mathop{\lim }}\,\frac{f\left( x \right)}{\sin
g\left( x \right)}=\frac{f\left( 1 \right)}{\sin g\left( 1
\right)}=\frac{0}{0}\,\,\,indform$
By L’Hospital Rule and Fundamental theorem of calculus we get :
$\overset{H.R}{\mathop{=}}\,\underset{x\to 1}{\mathop{\lim
}}\,\frac{\frac{d}{dx}f\left( x \right)}{\frac{d}{dx}\left( \sin g\left( x
\right) \right)}=\underset{x\to 1}{\mathop{\lim }}\,\frac{\sin \left(
{{x}^{2}}-1 \right)}{\frac{d}{dx}g\left( x \right)\cos g\left( x
\right)}=\underset{x\to 1}{\mathop{\lim }}\,\frac{\sin \left( {{x}^{2}}-1
\right)}{\left( {{x}^{2}}-1 \right)\cos \left( \int_{1}^{x}{\left( {{t}^{2}}-1
\right)dt} \right)}$
Let $u={{x}^{2}}-1\Rightarrow x=\sqrt{u+1}$ as $x\to 1\,\,\,,\,\,u\to
0$
So \[\underset{x\to 1}{\mathop{\lim }}\,\frac{\sin \left( {{x}^{2}}-1
\right)}{\left( {{x}^{2}}-1 \right)\cos \left( \int_{1}^{x}{\left( {{t}^{2}}-1
\right)dt} \right)}=\underset{u\to 0}{\mathop{\lim }}\,\frac{\sin u}{u\cos
\left( \int_{1}^{x}{\left( {{t}^{2}}-1 \right)dt} \right)}\]
$=\underset{u\to 0}{\mathop{\lim }}\,\frac{\sin u}{u}\times
\frac{1}{\cos \int_{1}^{\sqrt{u+1}}{\left( {{t}^{2}}-1 \right)dt}}=\frac{1}{\cos
\left( 0 \right)}=1$
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