Calculate this limit - my algbra isn't coming out

[jwpaine]

\(\displaystyle \L \lim_{x\to\sqrt{2}}\, \frac{x^{2}-2}{x - \sqrt{2}}\)

= \(\displaystyle \L \lim_{x\to\sqrt{2}}\, \frac{x^{2}-2}{x - \sqrt{2}}\cdot \frac{x+\sqrt{2}}{x+\sqrt{2}}\)

= \(\displaystyle \L \lim_{x\to\sqrt{2}}\, \frac{x^{3}+\sqrt{2}x^{2}-2x-2\sqrt{2}}{x^{2}-2}\)


= \(\displaystyle \L \lim_{x\to\sqrt{2}}\, \frac{x(x^2-2) + \sqrt{2}x^{2} -2\sqrt{2}}{x^{2}-2}\)

= \(\displaystyle \L \lim_{x\to\sqrt{2}}\, x + \sqrt{2}x^{2} -2\sqrt{2}\)

= \(\displaystyle \L \sqrt{2} + \sqrt{2}(\sqrt{2})^{2} -2\sqrt{2}\)

= \(\displaystyle \L \sqrt{2} + 2\sqrt{2} -2\sqrt{2}\)

= \(\displaystyle \L \sqrt{2}\)

But the real answer is \(\displaystyle \L 2\sqrt{2}\)

What did I mess up on?
Thanks!

 

[Loren]

What happens if you factor \(\displaystyle x^2 - 2\) into \(\displaystyle (x + \sqrt{2})(x - \sqrt{2}\)\) right off the bat?

 

[soroban]

Hello, jwpaine!

You cancelled illegally . . .

\(\displaystyle \L \lim_{x\to\sqrt{2}}\, \frac{x^{2}-2}{x - \sqrt{2}}\)

\(\displaystyle \L = \;\lim_{x\to\sqrt{2}}\, \frac{x^{2}-2}{x - \sqrt{2}}\cdot \frac{x+\sqrt{2}}{x+\sqrt{2}}\)

\(\displaystyle \L =\;\lim_{x\to\sqrt{2}}\, \frac{x^{3}+\sqrt{2}x^{2}-2x-2\sqrt{2}}{x^{2}-2}\)

\(\displaystyle \L =\;\lim_{x\to\sqrt{2}}\, \frac{x(x^2-2) + \sqrt{2}x^{2} -2\sqrt{2}}{x^{2}-2}\)
. . . . . . . . . . . . . . . . . . . . . . .\(\displaystyle {\color{blue}\nwarrow}\)
\(\displaystyle \L = \;\lim_{x\to\sqrt{2}}\, x + \sqrt{2}x^{2} -2\sqrt{2}\;\;\)\(\displaystyle {\color{blue}\leftarrow}\;\;\)Here!

Factor completely: \(\displaystyle \L\:\lim_{x\to\sqrt{2}}\,\frac{x(x^2-2)\,+\,\sqrt{2}(x^2-2)}{x^2-2} \;=\;\lim_{x\to\sqrt{2}}\,\frac{(x^2-2)(x+\sqrt{2})}{x^2-2}\)


Now cancel: \(\displaystyle \L\:\lim_{x\to\sqrt{2}}\,\frac{\sout{(x^2-2)}(x+\sqrt{2})}{\sout{x^2-2}} \;=\;\lim_{x\to\sqrt{2}}\,\left(x\,+\,\sqrt{2}\right)\;\;\cdots\;\;\) etc.