Limit exercise asked in csb.gov.lb in 1995

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 \[\underset{x\to 1}{\mathop{\lim }}\,\frac{\sin \left( {{x}^{2}}-1 \right)}{\left( {{x}^{2}}-1 \right)\cos \left( \int_{1}^{x}{\left( {{t}^{2}}-1 \right)dt} \right)}=\underset{u\to 0}{\mathop{\lim }}\,\frac{\sin u}{u\cos \left( \int_{1}^{x}{\left( {{t}^{2}}-1 \right)dt} \right)}\]

$=\underset{u\to 0}{\mathop{\lim }}\,\frac{\sin u}{u}\times \frac{1}{\cos \int_{1}^{\sqrt{u+1}}{\left( {{t}^{2}}-1 \right)dt}}=\frac{1}{\cos \left( 0 \right)}=1$