Proving Lim_{x->4}: $\frac{x-4}{\sqrt{x}-2}=4$

[ciubba]

Homework Statement

Prove $$lim_{x->4}\frac{x-4}{\sqrt{x}-2}=4$$

Homework Equations

Epsilon\delta definition

The Attempt at a Solution

I can see that a direct evaluation at 4 leads to an indeterminate form, so:

$$\frac{x-4}{\sqrt{x}-2}*\frac{\sqrt{x}+2}{\sqrt{x}+2} \; \mbox{simplifies to} \; \sqrt{x}+2 \; \mbox{when} \; x \ne 4$$

Via epsilon\delta definition,
$$|\sqrt{x}+2-4|<\epsilon \; \mbox{whenever} \; |x-4|<\delta$$

Expanding the right side,

$$-\delta<x-4<\delta$$
$$-\delta+4<x<\delta+4$$
$$\sqrt{-\delta+4}<\sqrt{x}<\sqrt{\delta+4} \; \mbox{when} \; \delta \leq 4$$
$$\sqrt{-\delta+4}-2<\sqrt{x}-2<\sqrt{\delta+4}-2$$

Thus, $$\delta=min\{\sqrt{-\delta+4}-2,4,\sqrt{\delta+4}-2\}$$

Given this definition of delta, it is elementary to work backwards towards $$|\sqrt{x}+2|<\delta$$

Have I made any errors in my proof?

 

[perplexabot]

Hey cuibba. I can't comment on your technique, but you could have used L'hopital's rule. That way I think is easier.

 

[pasmith]

Homework Statement

Prove $$lim_{x->4}\frac{x-4}{\sqrt{x}-2}=4$$

Homework Equations

Epsilon\delta definition

The Attempt at a Solution

I can see that a direct evaluation at 4 leads to an indeterminate form, so:

$$\frac{x-4}{\sqrt{x}-2}*\frac{\sqrt{x}+2}{\sqrt{x}+2} \; \mbox{simplifies to} \; \sqrt{x}+2 \; \mbox{when} \; x \ne 4$$

Via epsilon\delta definition,
$$|\sqrt{x}+2-4|<\epsilon \; \mbox{whenever} \; |x-4|<\delta$$Expanding the right side,

$$-\delta<x-4<\delta$$
$$-\delta+4<x<\delta+4$$
$$\sqrt{-\delta+4}<\sqrt{x}<\sqrt{\delta+4} \; \mbox{when} \; \delta \leq 4$$
$$\sqrt{-\delta+4}-2<\sqrt{x}-2<\sqrt{\delta+4}-2$$
Thus, $$\delta=min\{\sqrt{-\delta+4}-2,4,\sqrt{\delta+4}-2\}$$

Given this definition of delta, it is elementary to work backwards towards $$|\sqrt{x}+2|<\delta$$

Have I made any errors in my proof?

What is $\delta$ when $\epsilon = 10$? What about when $\epsilon = 10^{-5}$?

Your task is to prove:

For every $\epsilon > 0$ there exists a $\delta > 0$ such that if $0 < |x - 4| < \delta$ then $|\frac{x - 4}{\sqrt{x} - 2} - 4| < \epsilon$.

You need to relate your $\delta$ to the $\epsilon$.

 

[Ray Vickson]

Hey cuibba. I can't comment on your technique, but you could have used L'hopital's rule. That way I think is easier.

He could have used l'Hospital's rule (yes, it really is spelled Hospital) but his way is easier. Alternatively, he could have set $\sqrt{x} = y$ and then evaluated
$$\lim_{y \to 2} \frac{y^2 - 4}{y-2},$$
because in this form l'Hospital's rule is much nicer to use.

 

[ciubba]

Hey cuibba. I can't comment on your technique, but you could have used L'hopital's rule. That way I think is easier.

Ah, I wish I was allowed to use that technique, but our prof. is adamant on teaching delta ep. definitions first.

What is $\delta$ when $\epsilon = 10$? What about when $\epsilon = 10^{-5}$?

Your task is to prove:

For every $\epsilon > 0$ there exists a $\delta > 0$ such that if $0 < |x - 4| < \delta$ then $|\frac{x - 4}{\sqrt{x} - 2} - 4| < \epsilon$.

You need to relate your $\delta$ to the $\epsilon$.

Oops, I mistyped. I meant to say "Given this definition of delta, it is elementary to work backwards towards $$|\sqrt{x}+2\mathbf{-4}|<\boldsymbol{\epsilon}$$ My main question is whether or not it is correct to say that, for any epsilon, delta is $$min\{\sqrt{-\delta+4}-2,4,\sqrt{\delta+4}-2\}$$

He could have used l'Hospital's rule (yes, it really is spelled Hospital) but his way is easier. Alternatively, he could have set $\sqrt{x} = y$ and then evaluated
$$\lim_{y \to 2} \frac{y^2 - 4}{y-2},$$
because in this form l'Hospital's rule is much nicer to use.

Ah, I didn't think to do a u-sub. I'll definitely try that next time!