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Evaluate the following limits:

(a) $\lim _{t \rightarrow-3} \frac{t^{2}-9}{2 t^{2}+7 t+3}$

(b) $\lim _{h \rightarrow 0} \frac{(2+h)^{3}-8}{h}$

(a) $\lim _{t \rightarrow-3} \frac{t^{2}-9}{2 t^{2}+7 t+3}$

$=\frac{(-3)^{2}-9}{2(-3)^{2}+7(-3)+3}=\frac{0}{0}$

$\lim _{t \rightarrow -3} \frac{(t + 3)(t-3)}{(2 t+1)(t + 3)}=\lim _{t \rightarrow -3} \frac{t-3}{2 t+1}=\frac{-3-3}{2(-3)+1}=\frac{-6}{-5}=\frac{6}{5}$

(b) $\lim _{h \rightarrow 0} \frac{(2+h)^{3}-8}{h}=\frac{(2+0)^{3}-8}{0}=\frac{0}{0}$

$\lim _{h \rightarrow 0} \frac{(2+h)(2+h)(2+h)-8}{h}=\lim _{h \rightarrow 0} \frac{h^{3}+6 h^{2}+12 h+8-8}{h}$

$\lim _{h \rightarrow 0} \frac{h^{3}+6 h^{2}+12 h}{h}=\lim _{h \rightarrow 0} \frac{h\left(h^{2}+6 h+12\right)}{h}=(0)^{2}+6(0)+12=12$

Evaluate the following limits:

(a) $\lim _{x \rightarrow 3}(2 x+|x-3|)$

(b) $\lim _{t \rightarrow 1} \frac{t^{4}-1}{t^{3}-1}$

(a) $\lim _{x \rightarrow 3}(2 x+|x-3|)$

(1) $\lim_{x \rightarrow 3^{-}} (2 x+|x-3|)=\lim_{x \rightarrow 3^{-}} (2 x-x+3)=6-3+3= 6$

(2) $\lim_{x \rightarrow 3^{+}} (2 x+|x-3|)=\lim_{x \rightarrow 3^{+}} (2 x+x-3)=3 x-3=9-3=6$

$\lim_{x \rightarrow 3^{-}} (2 x+|x-3|)=\lim _{x \rightarrow 3^{+}}(2 x+|x-3|)=6$

(b) $\lim _{t \rightarrow 1} \frac{t^{4}-1}{t^{3}-1}=\frac{1-1}{1-1}=\frac{0}{0}$

$\lim _{t \rightarrow 1} \frac{\left(t^{2}+1\right)\left(t^{2}-1\right)}{(t-1)\left(t^{2}+t+1\right)}$

$=\lim _{t \rightarrow 1} \frac{\left(t^{2}+1\right)(t+1)(t-1)}{(t-1)\left(t^{2}+t+1\right)}$

$\lim _{t \rightarrow 1} \frac{\left(t^{2}+1\right)(t+1)}{\left(t^{2}+t+1\right)}=\frac{\left(t^{2}+1\right)(t+1)}{t^{2}+t+1}=\frac{(2)(2)}{1+1+1}=\frac{4}{3}$

Prove that $\lim _{x \rightarrow 0} x^{2} \cos \frac{2}{x}=0$

$\lim _{x \rightarrow 0} x^{2} \cos \frac{2}{x}=\quad \cos \frac{2}{6}=\cos (\infty)$

$-1 \leq \cos \frac{2}{x} \leq 1$

$\rightarrow-x^{2} \leq x^{2} \cos \frac{2}{x} \leq x^{2}$

$\lim _{x \rightarrow 0}\left(-x^{2}\right)=0 \quad \lim _{x \rightarrow 0}\left(x^{2}\right)=0$

by squeeze theotem $\longrightarrow \lim _{x \rightarrow 0} x^{2} \cos \frac{2}{x}=0$

Prove that $\lim _{x \rightarrow 0} \sqrt{x} \mathrm{e}^{\sin \left(\frac{\pi}{x}\right)}=0$

$\lim _{x \rightarrow 0} \sqrt{x} \mathrm{e}^{\sin \left(\frac{\pi}{x}\right)}=0 = \quad \sin \pi / 0$

$e^{-1} \leq e^{\sin \frac{\pi}{x}} \leq e^{1}$

$\sqrt{x} \: e^{-1} \leq \sqrt{x} \: e^{\sin \frac{\pi}{x}} \leq \sqrt{x} \: e^{1}$

$\lim _{x \rightarrow 0} \sqrt{x}\: e^{-1}=0$

$\lim _{x \rightarrow 0} \sqrt{x} \: e=0$

$\lim _{x \rightarrow 0} \sqrt{x} \: e^{\sin \pi / x}=0$