91Ó°ÊÓ

Determine values of \(a\) and \(b\) that make the given function continuous. $$f(x)=\left\\{\begin{array}{ll} \frac{2 \sin x}{x} & \text { if } x<0 \\ a & \text { if } x=0 \\ b \cos x & \text { if } x>0 \end{array}\right.$$

Evaluate the indicated limit, if it exists. Assume that \(\lim _{x \rightarrow 0} \frac{\sin x}{x}=1\) $$\lim _{t \rightarrow-2} \frac{\frac{1}{2}+\frac{1}{t}}{2+t}$$

Find an \(M\) or \(N\) corresponding to \(\varepsilon=0.1\) for each limit at infinity. $$\lim _{x \rightarrow-\infty} \frac{3 x^{2}-2}{x^{2}+1}=3$$

Determine values of \(a\) and \(b\) that make the given function continuous. $$f(x)=\left\\{\begin{array}{ll} a e^{x}+1 & \text { if } x<0 \\ \sin ^{-1} \frac{x}{2} & \text { if } 0 \leq x \leq 2 \\ x^{2}-x+b & \text { if } x>2 \end{array}\right.$$

Determine all horizontal and vertical asymptotes. For each vertical asymptote, determine whether \(f(x) \rightarrow \infty\) or \(f(x) \rightarrow-\infty\) on either side of the asymptote. $$f(x)=3 e^{-1 / x}$$

As we see in Chapter 2, the velocity of an object that has traveled \(\sqrt{x}\) miles in \(x\) hours at the \(x=1\) hour mark is given by \(v=\lim _{x \rightarrow 1} \frac{\sqrt{x}-1}{x-1} .\) Estimate this limit.

Evaluate the indicated limit, if it exists. Assume that \(\lim _{x \rightarrow 0} \frac{\sin x}{x}=1\) $$\lim _{x \rightarrow 0} \frac{\tan 2 x}{5 x}$$

Determine values of \(a\) and \(b\) that make the given function continuous. $$f(x)=\left\\{\begin{array}{ll} a\left(\tan ^{-1} x+2\right) & \text { if } x<0 \\ 2 e^{b x}+1 & \text { if } 0 \leq x \leq 3 \\ \ln (x-2)+x^{2} & \text { if } x>3 \end{array}\right.$$

Use numerical and graphical evidence to conjecture the value of \(\lim _{x \rightarrow 0} x^{2} \sin (1 / x) .\) Use the Squeeze Theorem to prove that you are correct: identify the functions \(f\) and \(h\) show graphically that \(f(x) \leq x^{2} \sin (1 / x) \leq h(x)\) and justify $$\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} h(x)$$

Find an \(M\) or \(N\) corresponding to \(\varepsilon=0.1\) for each limit at infinity. $$\lim _{x \rightarrow \infty} e^{-2 x}=0$$