91Ó°ÊÓ

Find the limit \(L\). Then use the \(\varepsilon-\delta\) definition to prove that the limit is \(L\). $$ \lim _{x \rightarrow 2}(x+3) $$

Find the limit. \(\lim _{x \rightarrow-3^{-}} \frac{x^{2}+2 x-3}{x^{2}+x-6}\)

Find the limit \(L\). Then use the \(\varepsilon-\delta\) definition to prove that the limit is \(L\). $$ \lim _{x \rightarrow-3}(2 x+5) $$

Use the information to evaluate the limits. \(\lim _{x \rightarrow c} f(x)=\frac{3}{2}\) \(\lim _{x \rightarrow c} g(x)=\frac{1}{2}\) (a) \(\lim _{x \rightarrow c}[4 f(x)]\) (b) \(\lim _{x \rightarrow c}[f(x)+g(x)]\) (c) \(\lim _{x \rightarrow c}[f(x) g(x)]\) (d) \(\lim _{x \rightarrow c} \frac{f(x)}{g(x)}\)

Find the limit. \(\lim _{x \rightarrow(-1 / 2)^{+}} \frac{6 x^{2}+x-1}{4 x^{2}-4 x-3}\)

Find the \(x\) -values (if any) at which \(f\) is not continuous. Which of the discontinuities are removable? $$ f(x)=\frac{x}{x^{2}-1} $$

Find the limit \(L\). Then use the \(\varepsilon-\delta\) definition to prove that the limit is \(L\). $$ \lim _{x \rightarrow-4}\left(\frac{1}{2} x-1\right) $$

Find the \(x\) -values (if any) at which \(f\) is not continuous. Which of the discontinuities are removable? $$ f(x)=\frac{x}{x^{2}+1} $$

Use the information to evaluate the limits. \(\lim _{x \rightarrow c} f(x)=4\) (a) \(\lim _{x \rightarrow c}[f(x)]^{3}\) (b) \(\lim _{x \rightarrow c} \sqrt{f(x)}\) (c) \(\lim _{x \rightarrow c}[3 f(x)]\) (d) \(\lim _{x \rightarrow c}[f(x)]^{3 / 2}\)

Find the limit. \(\lim _{x \rightarrow 1} \frac{x^{2}-x}{\left(x^{2}+1\right)(x-1)}\)