Prove $\lim_{x \to 2} \frac{1}{x} = \frac{1}{2}$ using delta epsilon definition

Question

Prove $\lim_{x \to 2} \frac{1}{x} = \frac{1}{2}$ using delta epsilon definition.

So I did $|x-2|<\delta$ and $|1/x -1/2|<\varepsilon$.

The second inequality can be written as $|(2-x)/2x|<\varepsilon$ and this can be written as $|(x-2)/2x|<\varepsilon$ but now there is a $2x$ in the denominator.

Because I could write $|x-2|<\varepsilon 2x$ but then delta depends on x if I choose $\delta= \varepsilon 2x$. I could take a nearby point and say $|x-2|<1$ so $1<x<3$ then delta would be $6\varepsilon$ this also doesnt seem right.

Answer 1

We need to show that given any $\varepsilon>0$ that there exists a number $\delta>0$ such that whenever $0<|x-2|<\delta$, $\left|\frac1x-\frac12\right|<\varepsilon$.

Since we are interested in the behavior of $\frac1x$ when $x$ is "close" to $2$, let's limit the interval over which we examine such behavior. So, let's agree to look in the interval $1<x<3$ or $|x-2|<1$. While this is arbitrary, it is fit for purpose here.

Now that we've agreed to restrict $x$ to the interval $(1,3)$, we can write

$$\left|\frac1x-\frac12\right|=\frac{|x-2|}{|2x|}<\frac12 |x-2|\tag1$$

The right-hand side of $(1)$ is less than and given $\varepsilon>0$ when $|x-2|<2\varepsilon$.

Hence, if we take $\delta>0$ as the smaller of $1$ and $2\varepsilon$, then we can assert that for any $\varepsilon>0$, there exists a $\delta=\min(1,2\varepsilon)$ such that

$$\left|\frac1x-\frac12\right|<\varepsilon$$

whenever $0<|x-2|<\delta$.

$$$$

Answer 2

$|x-2|< \delta \Leftrightarrow 2-\delta<x<2+\delta$ so if $\delta<1$ then $|x|=x>1$.

Now $|(2-x)/2x|=\frac{|2-x|}{2x}<\frac{|2-x|}{2}<\varepsilon$ can you finish from here?

Answer 3

If $x$ is a number such that $|x-2| < 1$, then $x>1$, and so $$\frac1{2x} < \frac12.$$ Hence, for any $x$ such that $0 < |x-2| < \min\{1,2\varepsilon\}$ we have that $$\bigg|\frac{x-2}{2x}\bigg| = \frac{1}{2x} \cdot |x-2| < \frac12 \cdot 2\varepsilon = \varepsilon.$$

Answer 4

Assuming wlog $|x-2|<1 \iff 1<x<3$ we have

$$\left|\frac{x-2}{2x}\right|<\frac{\left|x-2\right|}{2}<\varepsilon \iff|x-2|<2\varepsilon $$

then we can take $\delta =\min(1;2\varepsilon)$.

Answer 5

As you are looking for the limit when $ x $ goes to $ 2 $, You can assume that $ x $ is not far from $ 2 $. In other words, you can assume that $ x $ satisfies the condition : $$1<x<3\;\; \text{ or }\;\; |x-2|<\color{red}{1}$$

So $$|\frac{x-2}{2x}|<\frac 12|x-2|$$

thus, to satisfy the condition $$|\frac 1x-\frac 12|<\epsilon$$

we just need $$\frac 12|x-2|<\epsilon$$ or $$|x-2|<2\epsilon$$.

Finally, we look for $ \delta>0$ such that

$$|x-2|<\color{red}{1} \; and\; |x-2|<\delta \implies |x-2|<2\epsilon$$ Thus, we will choose $$\delta=\min(\color{red}{1},2\epsilon)$$