Proof verification: $\lim_{x\to2} \frac{1}{x} = \frac{1}{2}$

Question

Is either of the following methods correct?

Prove $$\lim_{x\to2} \frac{1}{x} = \frac{1}{2}$$

First Method

Preliminary Analysis:

We know that $a = 2$, $L= \frac{1}{2}$, and $f(x)= \frac{1}{x}$. By the precise definition of limit we have the following: $$ 0<|x-a|<\delta \implies |f(x)-L|<\varepsilon\\ 0<|x-2|<\delta \implies \left|\frac{1}{x}-\frac{1}{2}\right|<\varepsilon \implies \left|\frac{2}{2x}-\frac{x}{2x}\right| <\varepsilon \implies \left|\frac{2-x}{2x}\right| <\varepsilon \\ \implies \left|\frac{-(-2+x)}{2x}\right| <\varepsilon \implies \left|\frac{x-2}{2x}\right| <\varepsilon \implies \frac{\left|x-2\right|}{\left|2x\right|} <\varepsilon \implies \left|x-2\right|<\varepsilon \left|2x\right| $$

let $\delta = \varepsilon\left|2x\right|$ but we need to simplify it further because delta should be in terms of $\varepsilon$ only.

Assume $|x-a| < 1$

$$ |x-2| < 1 \implies -1 <x -2<1 \implies -1+2<x<1+2 \implies 1<x<3$$

Then we have to simplify $|2x|$ as well, which ends up being

$$ 2<2x<6 \implies -6<2x<6 \implies |2x| <6$$

Now consider the inequality we discovered, specifically, $\left|x-2\right|<\varepsilon \left|2x\right|$ this inequality is also valid for $\left|x-2\right|<\varepsilon\cdot6$.

Let $\delta = \min{\{1, \varepsilon\cdot6}\}$

Proof:

Given $\varepsilon > 0$ let $\delta = \min{\{1, \varepsilon\cdot6}\}$ if $ 0<|x-2|<\delta \implies |x-2|<1 \implies 1 <x < 3 \implies 2 < 2x < 6 \implies - 6 < 2x < 6 \\ \implies |2x| < 6$

We also have $|x - 2| < \varepsilon \cdot6$.

$$ \left|\frac{1}{x}-\frac{1}{2}\right|\implies \left|\frac{2}{2x}-\frac{x}{2x}\right| \implies \left|\frac{2-x}{2x}\right| \\ \implies \left|\frac{-(-2+x)}{2x}\right|\implies \left|\frac{x-2}{2x}\right| \implies \frac{\left|x-2\right|}{\left|2x\right|} < \frac{\varepsilon \cdot 6}{6} = \varepsilon $$

By the precise definition of limit $$\lim_{x\to2} \frac{1}{x} = \frac{1}{2}$$

Second Method

Proof by Definition/Property:

By the direct substitution property if $f$ is a polynomial or a rational function and $a$ is in the domain of $f$, then

$$\lim_{x \to a} f(x) = f(a)$$

Then by the direct substituon property of limit: $$\lim_{x\to2} \frac{1}{x} = \frac{1}{2}$$

Are either of the above methods correct? (I am putting the question here as well in case someone misses it)

Answer 1

For the second method, it is ok since the rational function is continuous on it definition area.

Answer 2

The first is incorrect.

$$|x-2|<1\implies 1 <x <3\implies $$ $$\frac {1}{6}<\frac {1}{2x}<\frac {1}{2} $$

thus

$$|\frac {1}{x}-\frac {1}{2}|<\frac {|x-2|}{2} $$

if $|x-2|<1$ and $|x-2|<2\epsilon $ then $$|\frac {1}{x}-\frac {1}{2}|<\epsilon $$

we take $$\delta=\min (1,2\epsilon) $$

Answer 3

proof-verification:

(1)Given $\varepsilon > 0$, let $\delta = \min{\{1, 6\varepsilon}\}$.

(2)if $ 0<|x-2|<\delta \implies |x-2|<1 \implies 1 <x < 3 \implies 2 < 2x < 6 \implies - 6 < 2x < 6 \\ \implies |2x| < 6$

(3)We also have $|x - 2| < \varepsilon \cdot6$.

$$ \left|\frac{1}{x}-\frac{1}{2}\right|\implies \left|\frac{2}{2x}-\frac{x}{2x}\right| \implies \left|\frac{2-x}{2x}\right| \\ \implies \left|\frac{-(-2+x)}{2x}\right|\implies \left|\frac{x-2}{2x}\right| \implies \frac{\left|x-2\right|}{\left|2x\right|} < \frac{\varepsilon \cdot 6}{6} = \varepsilon $$

The writing of these line 2 and 3 are terrible. One should not write those confusing big "implication" arrows for doing simple algebra. Moreover, (3) is incorrect: $|x-2|<6\varepsilon$ and $|2x|<6$ do not imply that $$ \frac{|x-2|}{|2x|}<\frac{6\varepsilon}{6}. $$

By the precise definition of limit, $$\lim_{x\to2} \frac{1}{x} = \frac{1}{2}$$

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You want to get the estimate like $$ \frac{|x-2|}{|2x|}<\epsilon. $$ The intuition is as follows. On the one hand, $|x-2|$ can be as small as possible if $x$ is close to $2$. One the other hand, when $x$ is close to $2$, the quantity $\frac{1}{|2x|}$ is bounded by some positive real number. Thus together, one can get $\frac{|x-2|}{|2x|}$ as small as one wants when $x$ is close to $2$.

Now we turn the intuition to a rigorous proof. Given $\epsilon>0$, let $\delta = \min\{1,\epsilon\}$. If $0<|x-2|<\delta$, then $1<x<3$, which implies that $$ \frac{1}{|2x|}<1 \tag{1} $$ On the other hand, by the definition of $\delta$, we also have $$ |x-2|<\epsilon.\tag{2} $$ Combining (1) and (2), we have the desired inequality $$ \frac{|x-2|}{|2x|}<\epsilon. $$