$\lim_{x \rightarrow 4} \frac{\sqrt{x}-2}{x-4}$. Did I choose my $\delta$ Properly?

Question

Preliminaries:

We know that, the limit as $ x \rightarrow a $ of a function $ f(x) $ is $ L$ if for every $ \epsilon > 0 $ there exist a $ \delta > 0 $ such that $ 0 <|x-a|<\delta \implies |f(x) - L|<\epsilon$. Our function is $ f(x) = \frac{\sqrt{x} - 2}{x - 4}$. Note that $$ \lim_{x \rightarrow 4} \frac{\sqrt{x} - 2}{x - 4} $$ cannot be solved as it is, for replacing $ x $ for $ 4 $ will render an undetermined answer. In order to fix this, we can rationalize the denominator by performing: $$ \frac{\sqrt{x} - 2}{x - 4} \cdot \frac{\sqrt{x}+2}{\sqrt{x}+2} = \frac{(x-4)}{(x-4)(\sqrt{x}+2)} = \frac{1}{\sqrt{x}+2} $$

Let the rationalized expression be $ g(x) $. as $ g(x) = f(x) $ we can compute the limit: $$ \lim_{x \rightarrow 4}\frac{1}{\sqrt{x}+2} = \frac{1}{\sqrt{4}+2} = \frac{1}{2+2} = \frac{1}{4} $$

The Process for Choosing the Delta:

We shall say, that the limit as $ x \rightarrow 4 $ of $ \frac{1}{\sqrt{x}+2}$ is $\frac{1}{4}$ if for every $ \epsilon > 0 $ there exist a $ \delta > 0$ such that $ 0 < | x - 4| < \delta \implies \left|\frac{1}{\sqrt{x}+2}-\frac{1}{4}\right|<\epsilon$

Thanks to the property of triangle inequality, we have that: $$ \left|\frac{1}{\sqrt{x}+2}-\frac{1}{4}\right| \leq \left| \frac{1}{\sqrt{x}+2}\right| - \left|\frac{1}{4}\right| $$ For instance, it is sufficient for $$ \left| \frac{1}{\sqrt{x}+2}\right| - \frac{1}{4} < \epsilon $$ which is equivalent to $$ \left|\frac{1}{\sqrt{x}+2}\right| < \frac{4\epsilon+1}{4} $$ Now, as per the definition of absolute value inequality, we have that:

$$ -\frac{4\epsilon + 1}{4} < \frac{1}{\sqrt{x}+2} < \frac{4\epsilon + 1}{4} $$

By raising the expression to the power of minus one on each "side of the inequality" we can argue that,

$$ -\frac{4}{4\epsilon +1} < \frac{\sqrt{x}+2}{1} < \frac{4}{4\epsilon+1} $$ which is equivalent to $$ \left|\sqrt{x}+2\right| < \frac{4}{4\epsilon+1} $$ by using the triangle inequality once again, we can argue that $\left|\sqrt{x}+2\right| \leq \left| \sqrt{x} \right| + |2|$

which means that it suffices for: $$ \left|\sqrt{x}\right| + 2 < \frac{4}{4\epsilon + 1} $$ $$ \left| \sqrt{x} \right| < \frac{4}{4\epsilon+1} - 2 $$ $$ \left|\sqrt{x}\right| < \frac{4-2(4\epsilon +1)}{4\epsilon+1} $$ If we square both sides of the equation, we shall have

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

Now, using once again the triangle inequality, we have that $$ \left|x-4\right| \leq \left|x\right| + \left|- 4\right| $$ for instance, if $$\left|x\right| + \left|- 4\right| < \left(\frac{4-2(4\epsilon +1)}{4\epsilon+1}\right)^2 + \left|-4\right| = \left(\frac{4-2(4\epsilon +1)}{4\epsilon+1}\right)^2 + 4$$ we will have $$ \left| x-4 \right| < \left(\frac{4-2(4\epsilon +1)}{4\epsilon+1}\right)^2 + 4$$

For instance, if we choose

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

The Limit will exist.

question: Did I choose my Delta Correctly, and for instance is my proof correct?

Answer 1

Thanks to the property of triangle inequality, we have that: $$ \left|\frac{1}{\sqrt{x}+2}-\frac{1}{4}\right| \leq \left| \frac{1}{\sqrt{x}+2}\right| \color{red}{\mathbf{-}} \left|\frac{1}{4}\right| $$

This is not correct.

Instead, you could rewrite: $$\left|\frac{1}{\sqrt{x}+2}-\frac{1}{4}\right|=\frac{1}{4}\left|\frac{2-\sqrt{x}}{2+\sqrt{x}}\right|=\frac{1}{4}\left|\frac{4-x}{\left(2+\sqrt{x}\right)^2}\right|=\frac{1}{4}\frac{\left|x-4\right|}{\left(2+\sqrt{x}\right)^2} \le \frac{\left|x-4\right|}{16}$$ And since $x \to 4$, you can make this arbitrarily small.