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• Notes

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

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

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

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

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

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