[SPLIT] lim [x -> 2] [ (x^2 - 4) / (x^2 - 2x) ]

[jwpaine]

Here's another one:

\(\displaystyle \L \lim_{x\to\2}\,\frac{x^{2}\,-\,4}{x^{2}\,-\,2x}\)

so to find the limit of the function as x approaches 2, I will let x grow by 2...

\(\displaystyle \L \frac{(x\,+\,2)^{2}\,-\,4}{(x\,+\,2)^{2}\,-\,2(x\,+\,2)}\, =\, \frac{x^{2}\,+\,4x}{x^{2}\,+\,2x}\, =\, 2\)

Correct?

Also: Can I view this as just finding the horizontal asymptote with the method we learned back in Algebra III? (Namely, this method?)

 

[galactus]

Here's another one:

\(\displaystyle \L \lim_{x\to\2}\,\,\frac{x^{2}\,-\,4}{x^{2}\,-\,2x}\)

Expand and get \(\displaystyle \L\\\lim_{x\to\2}\,\frac{2}{x}\,+\,1\,=\,2\)

so to find the limit of the function as x approaches 2, I will let x grow by 2...

\(\displaystyle \L \frac{(x\,+\,2)^{2}\,-\,4}{(x\,+\,2)^{2}\,-\,2(x\,+\,2)}\, =\, \frac{x^{2}\,+\,4x}{x^{2}\,+\,2x}\, =\, 2\)

No, not quite. This limit is 3/2, not 2.

\(\displaystyle \L\\\lim_{x\to\2}\, \frac{x^{2}\,+\,4x}{x^{2}\,+\,2x}\, =\,\lim_{x\to\2}\,\left[\frac{2}{2\,+\,x}\,+\,1\right]\, =\,\frac{3}{2}\)