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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:
$\lim_{x \to p} \frac{f(x)}{g(x)}=\frac{A}{B}$

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    $\begingroup$ 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 $\lim_{x \to 2}\frac{1}{x^2} = \frac{1}{4}$. $\endgroup$
    – Corrêa
    Mar 21 at 16:01
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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.

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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 $\frac{3}{2}<x<\frac{5}{2}$. 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.

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