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Find $$\lim _{x \rightarrow 1} \frac{\ln (x)}{x-1} \quad \lim _{x \rightarrow \infty} \frac{e^{x}}{x^{2}} \quad \lim _{x \rightarrow 2} \frac{x^{2}-4}{x-2}$$

$$\lim _{x \rightarrow 1} \frac{\ln (x)}{x-1}=\frac{\ln (1)}{1-1}=\frac{0}{0}$$

$$\lim _{x \rightarrow 1} \frac{\ln (x)}{x-1}=\lim _{x \rightarrow 1} \frac{1 / x}{1}=\lim _{x \rightarrow 1} 1 / x=\frac{1}{1}=1$$

$$\lim _{x \rightarrow \infty} \frac{e^{x}}{x^{2}}=\lim \frac{e^{\infty}}{(\infty)^{2}}=\frac{\infty}{\infty}$$

$$\lim _{x \rightarrow \infty} \frac{e^{x}}{x^{2}}=\lim _{x \rightarrow \infty} \frac{e^{x}}{2 x}=\frac{\infty}{\infty}=\lim _{x \rightarrow \infty} \frac{e^{x}}{2}=\frac{e^{\infty}}{2}=\infty$$

$$\lim _{x \rightarrow 2} \frac{x^{2}-4}{x-2}=\frac{(2)^{2}-4}{2-2}=\frac{0}{0} \:,\: \lim _{x \rightarrow 2} \frac{x^{2}-4}{x-2}=\lim _{x \rightarrow 2}\frac {2x}{1}= 2(2)=4$$

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