What is the limit $\lim_{x\to 3}\frac{1}{(3-x)^2}$?

Question

Why does the limit of $\frac{1}{(3-x)^2}$ become infinity as $x\to 3$?

When I simplify the expression I get $$ \frac{1}{(9-6x+x^2)}, $$ which would give the limit $-\frac{1}{18}$ as $x$ approaches $3$. I don't understand what goes wrong.

Answer 1

Notice that the term is positive all the time, we can't possibly have negative limit.

Notice that as $x \to 3$, $(x-3)^2$ approaches $0$.

Answer 2

Small numbers have large reciprocals, and $x-3$ gets small as $x \rightarrow 3.$ The square of $x-3$ is positive (and small) for $x$ near 3, so its reciprocal is positive (and large). The value of $\frac{1}{x-3}$ tends to plus infinity from either side of 3, so the limit does not exist (or, is $+\infty,$ depending on the conventions in place).

Answer 3

You shouldn't be considering what the fraction equates to at x=3, but rather what happens when x approaches 3 from the left and right side. What I mean by this is, consider x coming in from the negative and positive sides of the horizontal axis.

First, notice when x = 3, the fraction becomes 1/0, not -1/18. So let us consider the limit of the fraction as x approaches 3.

When x approaches 3 from the negative side, we can get very close to 3, say 2.999999.

Similarly, if we come in from the positive side, we can get close to 3, say 3.000000001.

When you plug these numbers back into the equation, you can see that the fraction becomes very positively large in both cases.

However it can get larger if we extended the decimals, therefore the Limit as x approaches 3 of the fraction becomes positive infinity.

Answer 4

Let $y=x-3 \to 0$, then

$$\lim_{x\to 3}\frac{1}{(3-x)^2}=\lim_{y\to 0}\frac{1}{y^2}$$

then note that for any $M>0$

$$\frac{1}{y^2}\ge M \implies |y|\le \frac1{\sqrt M}$$

therefore by definition the limit is $\infty$.