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


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)

  • $\begingroup$ What is your question? $\endgroup$ – Vincent Sep 27 '17 at 17:51
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    $\begingroup$ Use $\to$ for $\to$. $\endgroup$ – Shaun Sep 27 '17 at 17:51
  • $\begingroup$ @Vincent Are either of the following methods correct? It is also indicated at the very top. $\endgroup$ – Iamlearningmath Sep 27 '17 at 17:52
  • $\begingroup$ Look at the first sentence, @Vincent. $\endgroup$ – Shaun Sep 27 '17 at 17:52
  • $\begingroup$ The first proof is a little bit overkill in the level of detail between the steps, but is nonetheless accurate. The second proof, needs one small detail....The statement is true for rational functions when the denominator is non-zero. $\endgroup$ – Doug M Sep 27 '17 at 17:54


(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, $$\lim_{x\to2} \frac{1}{x} = \frac{1}{2}$$

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 $$ \frac{|x-2|}{|2x|}<\epsilon. $$


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


The first is incorrect.

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


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

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

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

  • $\begingroup$ So let me link the document that I used as a learning tool to come up with the first proof and perhaps you can point what I missed? Its the 4th proof. ocf.berkeley.edu/~yosenl/math/epsilon-delta.pdf $\endgroup$ – Iamlearningmath Sep 27 '17 at 18:32
  • $\begingroup$ @Iamlearningmath Your case is not in the document. $\endgroup$ – hamam_Abdallah Sep 27 '17 at 21:00

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