WeBWorK Standalone Renderer

Put the following statements in order to justify why $\displaystyle{\lim_{n \to \infty} \frac{{2n-2+5n^{2}}}{{7n^{2}+5n+1}} = \frac{5}{7}}$ .

Statements to choose from: Drag these statements to the right column.

$= \displaystyle \lim_{n \to \infty} \frac{\displaystyle n^2 \left( 5+\frac{2}{n}-\frac{2}{n^{2}}\right) }{\displaystyle n^2 \left(7+\frac{5}{n}+\frac{1}{n^{2}}\right) }$ .
$= \displaystyle \frac{\displaystyle \lim_{n \to \infty} \left( 5 \right) +2 \lim_{n \to \infty} \left( \frac{1}{n} \right) -2 \lim_{n \to \infty} \left( \frac{1}{n^2} \right) }{\displaystyle \lim_{n \to \infty} \left( 7 \right) +5 \lim_{n \to \infty} \left( \frac{1}{n} \right) +1 \lim_{n \to \infty} \left( \frac{1}{n^2} \right) }$
$= \displaystyle \lim_{n \to \infty} \frac{\displaystyle 5+\frac{2}{n}-\frac{2}{n^{2}}}{\displaystyle 7+\frac{5}{n}+\frac{1}{n^{2}}}$ .
$= \displaystyle \frac{5 +2 \cdot 0 -2 \cdot 0}{7 +5 \cdot 0 +1 \cdot 0}$
$= \displaystyle \frac{5}{7}.$
$= \displaystyle \frac{\displaystyle \lim_{n \to \infty} \left( 5+\frac{2}{n}-\frac{2}{n^{2}} \right)}{ \displaystyle \lim_{n \to \infty} \left( 7+\frac{5}{n}+\frac{1}{n^{2}} \right)}$ .
$\displaystyle \lim_{n \to \infty} \frac{2n-2+5n^{2}}{7n^{2}+5n+1}$ .

Your justification: Put the statements in order in this column and press the Submit Answers button.

Statements to choose from: Drag these statements to the right column.

$= \displaystyle \lim_{n \to \infty} \frac{\displaystyle n^2 \left( 5+\frac{2}{n}-\frac{2}{n^{2}}\right) }{\displaystyle n^2 \left(7+\frac{5}{n}+\frac{1}{n^{2}}\right) }$ .
$= \displaystyle \frac{\displaystyle \lim_{n \to \infty} \left( 5 \right) +2 \lim_{n \to \infty} \left( \frac{1}{n} \right) -2 \lim_{n \to \infty} \left( \frac{1}{n^2} \right) }{\displaystyle \lim_{n \to \infty} \left( 7 \right) +5 \lim_{n \to \infty} \left( \frac{1}{n} \right) +1 \lim_{n \to \infty} \left( \frac{1}{n^2} \right) }$
$= \displaystyle \lim_{n \to \infty} \frac{\displaystyle 5+\frac{2}{n}-\frac{2}{n^{2}}}{\displaystyle 7+\frac{5}{n}+\frac{1}{n^{2}}}$ .
$= \displaystyle \frac{5 +2 \cdot 0 -2 \cdot 0}{7 +5 \cdot 0 +1 \cdot 0}$
$= \displaystyle \frac{5}{7}.$
$= \displaystyle \frac{\displaystyle \lim_{n \to \infty} \left( 5+\frac{2}{n}-\frac{2}{n^{2}} \right)}{ \displaystyle \lim_{n \to \infty} \left( 7+\frac{5}{n}+\frac{1}{n^{2}} \right)}$ .
$\displaystyle \lim_{n \to \infty} \frac{2n-2+5n^{2}}{7n^{2}+5n+1}$ .

Your justification: Put the statements in order in this column and press the Submit Answers button.