Question

1. find the limit x approaches 0, f(x)=x/|x|
2. find the limit x approaches 2,
f(x)=|x-2|/x^2+x-6

Answer 1

\[|x|=\begin{cases}
x&\text{for }x\ge0\\-x&\text{for }x<0
\end{cases}\]
which means
\[|x-2|=\begin{cases}
x-2&\text{for }x\ge0\\-(x-2)&\text{for }x<0
\end{cases}\]
So for the first limit, when \(x\to0\), consider what happens when \(x\to0\) from either side. That is, consider the one-sided limits
\[\lim_{x\to0^-}f(x)~~~~\text{and}~~~~\lim_{x\to0^+}f(x)\]
If \(x\to0^-\) (approaching from the left), then that means any value of \(x\) would be negative, i.e. \(x<0\). This would mean \(|x|=-x\), so \(f(x)=\dfrac{|x|}{x}=\dfrac{-x}{x}=-1\). The limit would then be -1.

From the other side, if \(x\to0^+\), then \(x>0\), so \(|x|=x\) and \(f(x)=\dfrac{|x|}{x}=\dfrac{x}{x}=1\), and so the limit is 1.

The one-sided limits are not equivalent, so \(\displaystyle\lim_{x\to0}\frac{|x|}{x}\) does not exist.

Answer 2

Use this as an example to reason through the second question.

Answer 3

in the second question , the absolute value is the numerator. can i still use the same approach?

Answer 4

You can. The denominator can be factored.

Answer 5

\[\frac{ \left| x -2 \right| }{ \left( x +3 \right)\left( x -2 \right) }\]
\[=\frac{ 1 }{ x +3 }\]
from the left,\[\frac{ 1 }{ -\left( 2+3 \right) }=\frac {- 1}{5}\]
and the right will be \[\frac{ 1 }{ 5 }\]
is that it?