Limit exercises asked by 3 Arabic teachers in the mathematics teacher group

Exercise:

Compute, $\underset{x\to e}{\mathop{\lim }}\,\frac{\sin x-\sin e}{\ln x-1}$

Solution: we have $\frac{\sin x-\sin e}{\ln x-1}=\frac{\sin x-\sin e}{\ln x-\ln e}\times \frac{x-e}{x-e}=\frac{\sin x-\sin e}{x-e}\times \frac{x-e}{\ln x-\ln e}$

So $\underset{x\to e}{\mathop{\lim }}\,\frac{\sin x-\sin e}{\ln x-\ln 1}=\underset{x\to e}{\mathop{\lim }}\,\frac{\sin x-\sin e}{x-e}\times \underset{x\to e}{\mathop{\lim }}\,\frac{x-e}{\ln x-\ln e}=\frac{d}{dx}{{\left( \sin x \right)}_{x=e}}\frac{1}{\frac{d}{dx}{{\left( \ln x \right)}_{x=e}}}$

Thus $\underset{x\to e}{\mathop{\lim }}\,\frac{\sin x-\sin e}{\ln x-1}=\left( \cos e \right)\frac{1}{\frac{1}{e}}=e\cos e$

Exercise:

Compute, $\underset{x\to 0}{\mathop{\lim }}\,\frac{2-\cos 2x-\cos 4x}{{{x}^{2}}}$

Solution: we have $\frac{2-\cos 2x-\cos 4x}{{{x}^{2}}}=\frac{1-\cos 2x+1-\cos 4x}{{{x}^{2}}}=\frac{2{{\sin }^{2}}x}{{{x}^{2}}}+\frac{2{{\sin }^{2}}2x}{{{x}^{2}}}$

So $\underset{x\to 0}{\mathop{\lim }}\,\frac{2-\cos 2x-\cos 4x}{{{x}^{2}}}=\underset{x\to 0}{\mathop{\lim }}\,\frac{2{{\sin }^{2}}x}{{{x}^{2}}}+\underset{x\to 0}{\mathop{\lim }}\,\frac{2{{\sin }^{2}}2x}{{{x}^{2}}}$

$$=2\underset{x\to 0}{\mathop{\lim }}\,{{\left( \frac{\sin x}{x} \right)}^{2}}+2\underset{x\to 0}{\mathop{\lim }}\,{{\left( \frac{\sin 2x}{x} \right)}^{2}}=2+2\underset{x\to 0}{\mathop{\lim }}\,\frac{{{\sin }^{2}}2x}{{{\left( \frac{2x}{2} \right)}^{2}}}=2+2\left( 4 \right)=10$$

Exercise:

Find the value of $a$ such that $\underset{x\to a}{\mathop{\lim }}\,\frac{{{x}^{2}}-{{a}^{2}}}{{{\left( x+a \right)}^{3}}-8{{a}^{3}}}=3$

Solution: we know that ${{x}^{2}}-{{a}^{2}}=\left( x-a \right)\left( x+a \right)$ and

${{\left( x+a \right)}^{3}}=\left( x+a \right)\left( {{x}^{2}}+2ax+{{a}^{2}} \right)$

So $\underset{x\to a}{\mathop{\lim }}\,\frac{{{x}^{2}}-{{a}^{2}}}{{{\left( x+a \right)}^{3}}-8{{a}^{3}}}=\underset{x\to a}{\mathop{\lim }}\,\frac{\left( x-a \right)\left( x+a \right)}{{{\left( x+a \right)}^{3}}-{{\left( 2a \right)}^{3}}}$

$=\underset{x\to a}{\mathop{\lim }}\,\frac{\left( x-a \right)\left( x+a \right)}{\left( x+a-2a \right)\left( {{\left( x+a \right)}^{2}}+2a\left( x+a \right)+4{{a}^{2}} \right)}=\underset{x\to a}{\mathop{\lim }}\,\frac{x+a}{{{\left( x+a \right)}^{2}}+2a\left( x+a \right)+4{{a}^{2}}}=3$

$\Leftrightarrow \frac{2a}{4{{a}^{2}}+2a\left( 2a \right)+4{{a}^{2}}}=3\Leftrightarrow \frac{2a}{12{{a}^{2}}}=3\Leftrightarrow \frac{1}{6a}=3\Leftrightarrow a=\frac{1}{18}$