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Find $$\lim _{x \rightarrow-3^{+}} \frac{x+2}{x+3}$$

$$x=-3 \quad \Rightarrow \lim _{x \rightarrow{-3}^+} \frac{x+2}{x+3}=\frac{-3+2}{-3+3}=\frac{-1}{0}=\pm \infty$$

$$x=-2 \cdot 9 \Rightarrow \lim _{x \rightarrow{-3}^{+ }}\frac{x+2}{x+3}=\frac{-2 \cdot 9+2}{-2 \cdot 9+3}=\frac{-0.9}{+0 \cdot 1}=-$$

$$\lim _{x \rightarrow{-3}^{+}} \frac{x+2}{x+3}=-\infty$$

Find $$\lim _{x \rightarrow 3^{+}} \ln \left(x^{2}-9\right)$$

$$\ln \left(3^{2}-9\right)=\ln (0)$$

$$(3.01)$$

$$\ln \left((3.01)^{2}-9\right)=\ln (+0.01)$$

$$\ln \left(3^{2}-9\right)=\ln (0)=-\infty$$

Find $$\lim _{x \rightarrow 2} \frac{x^{2}-x+6}{x-2}$$

$$\lim _{x \rightarrow 2} \frac{x^{2}-x+6}{x-2}=\frac{(2)^{2}-2+6}{2-2}=\frac{8}{0}=\pm \infty$$

 DNE

Find $$\lim _{x \rightarrow 16} \frac{4-\sqrt{x}}{16 x-x^{2}}$$

$$\lim _{x \rightarrow 16} \frac{4-\sqrt{x}}{16 x-x^{2}}=\frac{4-\sqrt{16}}{16(16)-16^{2}}=\frac{4-4}{16^{2}-16^{2}}=\frac{0}{0}$$

$$\lim _{x \rightarrow 16} \frac{4-\sqrt{x}}{x(16-x)}=\lim _{x \rightarrow 16} \frac{4- \sqrt{x}}{x(4+\sqrt{x})(4-\sqrt{x})}$$

$$=\lim _{x \rightarrow \infty} \frac{1}{x(4+\sqrt{x})}$$

$$=\frac{1}{16(4+\sqrt{16})}=\frac{1}{16(4+4)}=\frac{1}{16(8)}$$

$$=\frac{1}{128}$$

Find $$\lim _{t \rightarrow 0} \frac{\sqrt{1+t}-\sqrt{1-t}}{t}$$

$$\lim _{t \rightarrow 0} \frac{\sqrt{1+t}-\sqrt{1-t}}{t}=\frac{\sqrt{1}-\sqrt{1}}{0}=\frac{0}{0}$$

$$\lim _{t \rightarrow 0} \frac{(\sqrt{1+t}-\sqrt{1-t})(\sqrt{1+t}+\sqrt{1-t})}{t(\sqrt{1+t}+\sqrt{1-t})}$$

$$\lim_{t\rightarrow{0}}\frac{(1+t)-(1-t)}{t(\sqrt{1+t}+\sqrt{1-t})}$$

$$\lim _{t \rightarrow 0} \frac{2 t}{t(\sqrt{1+t}+\sqrt{1-t})}$$

$$=\frac{2}{\sqrt{1}+\sqrt{1}}=\frac{2}{2}=1$$

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