# computing the limit and stating the basic limit used $\lim_{x \to 2} \frac{1}{x^2}$

I'm trying to compute the following limit whilst using the basic limit theorems, although I'm slightly rusty and would appreciate you honest feedback.

$$\lim_{x \to 2} \frac{1}{x^2}$$

Given that when $$|x-2| < \delta$$

Then $$\frac{1}{x^2}=\frac{1}{4}$$

This is my reasoning thus, far, and from this I proceeded with:

$$\lim_{x \to 2} \frac{1}{x^2} = \frac{1}{4}$$

$$|\frac{1}{x^2}-\frac{1}{4}|<\epsilon$$

Though i'm not sure where to go next.As for the basic limit theorems, I tried approaching this with:
$$\lim_{x \to p} \frac{f(x)}{g(x)}=\frac{A}{B}$$

• If you can show that $\lim_{x \to 2} x = 2$ by definition, then using properties of limits you can conclude that $\lim_{x \to 2} x^2 = 4$ and so, since the limit is nonzero, you can conclude that $\lim_{x \to 2}\frac{1}{x^2} = \frac{1}{4}$. Mar 21 at 16:01

You seem to be incorporating some version of an $$\epsilon$$-$$\delta$$ argument, which is unnecessary if you're using basic limit theorems. For the record though, your statement "Given that when $$|x-2|<\delta$$ then $$\frac{1}{x^2}=\frac{1}{4}$$" is a misrepresentation of $$\epsilon$$-$$\delta$$ arguments. It would be $$|\frac{1}{x^2}-\frac{1}{4}|<\epsilon$$, as you have later down. With regards to that portion, though "I proceeded with: $$\lim_{x\to 1}\frac{1}{x^2}=\frac{1}{4}$$, $$|\frac{1}{x^2}-\frac{1}{4}|<\epsilon$$" is a bit of a reversal of the chain of reasoning you would present in an $$\epsilon$$-$$\delta$$ argument.

Anyhow, I just wanted to point those things out, but since this is not supposed to be an $$\epsilon$$-$$\delta$$ argument, I'll leave it there. As for the question at hand, Correa is correct in his comment.

Let $$f(x)=1$$ and let $$g(x)=x$$. You can use the property you stated, namely $$\lim_{x\to p}\frac{f(x)}{g(x)}=\frac{f(p)}{g(p)}$$ (provided each limit exists separately and $$g(p)\neq 0$$), but you must first establish that $$\lim_{x\to 2}x=2$$. You can then use $$\lim_{x\to p}f(x)g(x)=\lim_{x\to p}f(x)\lim_{x\to p}g(x)$$ to get the final answer.

The definition for limit of a function at a point is as follows:

If for all $$\epsilon>0$$, there exists $$\delta>0$$ such that $$0<|x-x_0|<\delta$$ implies $$|f(x)-L|<\epsilon$$, then the limit of $$f(x)$$ as $$x$$ approaches $$x_0$$ exists and is equal to $$L$$.

Here is a correct version of the $$\epsilon,\delta$$ argument using the definition:

Let $$\epsilon>0$$ be given and define $$\delta=\min\left\{\epsilon,\frac{1}{2}\right\}$$. Then $$|x-2|<\delta$$ implies $$\frac{3}{2}. However, we also have that

$$\left|\frac{1}{x^2}-\frac{1}{4}\right|=\frac{|4-x^2|}{4x^2}=\frac{|2-x||2+x|}{4x^2}=\frac{|x-2|(x+2)}{4x^2}$$

and therefore

$$<\frac{\delta(5/2+2)}{4\cdot (3/2)^2}=\frac{1}{2}\delta$$

Now, if $$\delta=\epsilon$$ then we are done as

$$\left|\frac{1}{x^2}-\frac{1}{4}\right|<\frac{1}{2}\delta=\frac{1}{2}\epsilon<\epsilon$$

If not, then $$\delta=\frac{1}{2}<\epsilon$$ and therefore

$$\left|\frac{1}{x^2}-\frac{1}{4}\right|<\frac{1}{2}\delta=\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}<\frac{1}{2}<\epsilon$$

and we are done.