Does following function have a limit if x approaches 2. Calculate what the limit is and motivate why if it is missing. $$ \frac{(x-2)^2}{(x-2)^3} =\frac{ 1 }{ x-2}. $$ I answered $\frac{1 }{ 0 }= 0 $ undefined is that correct?
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It looks like you are considering the function $$ f(x) = \frac{(x-2)^2}{(x-2)^3} = \frac{1}{x-2}. $$ You want to consider what happens to this function when $x$ approaches $2$. Note that the numerator is just the constant $1$ and when $x$ approaches $2$, then $x - 2$ approaches $0$. So you have something that approaches $1$ divided by $0$. This limit does not exist (as you correctly state).
Note, however, that $1$ divided by $0$ is not equal to $0$. In fact $1$ divided by $0$ is undefined which is the reason that the limit is undefined.
If you consider the limit as $x$ approaches $0$ from the right, then you are just considering what happens to $1 / x$ for positive values of $x$. And since you are taking this (non-zero) constant and dividing it by something that becomes smaller and smaller (while being positive), then the limit is $\infty$: $$ \lim_{x\to 0^+} f(x) = \infty. $$ Likewise $$ \lim_{x\to 0^-} f(x) = -\infty. $$
Thomas
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The denominator approaches zero and the numerator doesn't, so the limit "does not exist". If you've seen the term "indeterminate", then the reason is that 1/0 is undefined rather than indeterminate. However it's important to note that 1/0 is not 0, it's undefined.
Mark S.
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