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Evaluate the following limits.

1.   While it's true that $$\lim_{x \rightarrow 0} \frac{\sin 2x}{3x} = \frac{2}{3},$$ what's wrong with the following "proof"? $$\begin{array}{ll} & \lim_{x \rightarrow 0} \frac{\sin 2x}{3x} \\ = & \lim_{x \rightarrow 0} \frac{2\sin x}{3x} \\ = & \lim_{x \rightarrow 0} \frac{2}{3} \frac{\sin x}{x} \\ = & \frac{2}{3}\lim_{x \rightarrow 0} \frac{\sin x}{x} \\ = & \frac{2}{3} (1) \\ = & \frac{2}{3}\end{array}$$

a.  \(\sin 2x \neq 2\sin x\)

b.  You can't move the \(\frac{2}{3}\) from inside the limit to outside.

c.  \(\lim_{x\rightarrow 0} \frac{\sin x}{x} \neq 1\)

c.  Nothing is wrong here.

2.   Consider the graph in section 1.4 question 4. At \( x = -2 \)

a.  the function has a removable discontinuity.

b.  the function has a jump discontinuity.

c.  the function has an infinite discontinuity

d.  the function has an oscillatory discontinuity.

e.  the function does not have a discontinuity.

3.   Consider the graph in section 1.4 question 6. At \( x = -1 \)

a.  the function has a removable discontinuity.

b.  the function has a jump discontinuity.

c.  the function has an infinite discontinuity

d.  the function has an oscillatory discontinuity.

e.  the function does not have a discontinuity.

4.   If we wanted to use the squeeze theorem to show that the limit $$\lim_{x \rightarrow 0} x \sin(\frac{1}{x}) = 0,$$ we could sandwich the function \(f(x) = x \sin(\frac{1}{x})\) between which two functions?

a.  \(h(x) = -x^2\) and \(g(x) = x^2\)

b.  \(h(x) = -\mid x \mid\) and \(g(x) = \mid x \mid\)

c.  \(h(x) = -x^3\) and \(g(x) = x^3\)

d.  \(h(x) = -x^4\) and \(g(x) = x^4\)