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Order 7 of the following sentences so that they form a logical direct proof of the statement:

$\displaystyle{\lim_{x \rightarrow 4} (x^2) = 16}.$

Choose from these sentences:

Assume $0<|x-4|<\delta$
Assume $|x-4|<\delta$ and $| (x^2) - 16 | \leq \epsilon$
There is no limit.
$|x-4| | x+4 | < \frac{\epsilon}{10}(10) = \epsilon$
Choose $\delta>0$ so that $\delta<\frac{\epsilon}{10}$ and $\delta< 2$ .
Then $|x-4|<\delta$
$\delta < 2 \implies \\ 0 < x+4 < 8 + \delta < 8 + 2$
Therefore, $\displaystyle{\lim_{x \rightarrow 4} (x^2) = 16}.$
$8 - \delta < x+4 < 8 + \delta$
Use Wolfram-Alpha to figure this out.
Assume $| (x^2) - 16 | \leq \epsilon$ .
Suppose $\epsilon>0$

Your Proof:

Choose from these sentences:

Assume $0<|x-4|<\delta$
Assume $|x-4|<\delta$ and $| (x^2) - 16 | \leq \epsilon$
There is no limit.
$|x-4| | x+4 | < \frac{\epsilon}{10}(10) = \epsilon$
Choose $\delta>0$ so that $\delta<\frac{\epsilon}{10}$ and $\delta< 2$ .
Then $|x-4|<\delta$
$\delta < 2 \implies \\ 0 < x+4 < 8 + \delta < 8 + 2$
Therefore, $\displaystyle{\lim_{x \rightarrow 4} (x^2) = 16}.$
$8 - \delta < x+4 < 8 + \delta$
Use Wolfram-Alpha to figure this out.
Assume $| (x^2) - 16 | \leq \epsilon$ .
Suppose $\epsilon>0$

Your Proof: