Limit of a Piecewise Function

[Jason76]

\(\displaystyle g(x) = x^{2} - 1\), if \(\displaystyle x < 0\)

\(\displaystyle g(x) = 2x + 1\), if \(\displaystyle x\) is greater than or \(\displaystyle = 0\)

Find \(\displaystyle \lim\) of \(\displaystyle g(x)\) as \(\displaystyle x \rightarrow 0\)

Some starting hint?

 

[Jason76]

\(\displaystyle g(x) = (0)^{2} - 1 = -1\) Left hand limit \(\displaystyle \lim\) of \(\displaystyle g(x)\) as \(\displaystyle x \rightarrow 0-\)

\(\displaystyle g(x) = 2(0) + 1 = 1\), \(\displaystyle \lim\) of \(\displaystyle g(x)\) as \(\displaystyle x \rightarrow 0+\)

Limits don't match so \(\displaystyle \lim\) of \(\displaystyle g(x)\) as \(\displaystyle x \rightarrow 0\) DNE. Is that right?

 

[stapel]

\(\displaystyle g(x) = x^{2} - 1\), if \(\displaystyle x < 0\)

\(\displaystyle g(x) = 2x + 1\), if \(\displaystyle x\) is greater than or \(\displaystyle = 0\)

Find \(\displaystyle \lim\) of \(\displaystyle g(x)\) as \(\displaystyle x \rightarrow 0\)

Some starting hint?

Since the function is defined at the endpoints of each "half", you can simply do plug-n-chug. For the first rule, the value at zero is (or would be, according to the limit concept) equal to -1. For the second rule, the value at zero is +1. Since the "halves" don't "line up", the function can't be continuous.