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|<\deltaExpanding 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 <br /> |\sqrt{x}+2\mathbf{-4}|&lt;\boldsymbol{\epsilon}<br /> My main question is whether or not it is correct to say that, for any epsilon, delta is <br /> 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!