# How to evaluate a limit with roots

1. Sep 26, 2013

### MathewsMD

1. The problem statement, all variables and given/known data

I have the limit: lim [ (x+h)^2/3 - x^(2/3) /h ]

How would I further simplify and evaluate this limit.

2. The attempt at a solution

I have tried using a change of variable and using this in the sum of cubes formula (i.e. (x+h)^(2/3) = a, x^(2/3) = b, and then plugging this in to the sum of a cube equation. This did not help and just further complicated matters.

Next, I tried multiplying the numerator and denominator by {x+h)^(1/3) + x^(1/3) to cancel out the roots, but this just made new ones that were even harder to factor out.

I keep trying variations of the two methods I just described, but none of them work. Any direction or help on what to do next would be great. Thanks!

2. Sep 26, 2013

### Dick

Your sum of cubes formula really should help. (a-b)(a^2+ab+b^2)=a^3-b^3. Show why it doesn't? Multiply numerator and denominator by (a^2+ab+b^2).

3. Sep 26, 2013

### MathewsMD

Well I had h in the denominator, and what I was doing never allowed me to factor out the h in the expression. I ended up getting limh→0 [(a^(1/3) - b^(1/3))(a^2/3) + a^(1/3)b^(1/3) + b^(2/3) / h] and a = (x+h)^(2/3) and b = x^(2/3)

It got very messy, and I'm not sure if it's because I made an error in my steps, but I couldn't really go much further than this. I could just never factor out the h when I substituted the values for a and b back in. Any advice/thoughts would be appreciated.

4. Sep 27, 2013

### Simon Bridge

$$\lim_{h\rightarrow 0}\frac{(x+h)^{2/3}-x^{2/3}}{h}$$ ... is the derivative of $x^{2/3}$.

Is the problem to show this derivative by the definition or just to show the derivative?

if the latter then you can rewrite it as $y^3(x)=x^2$ and use the power rule.

If the former - OK: then you need to check your algebra if you cannot get h to factor out.
Just looking at it - for some reason you ended up with h alone in the denominator ... don't do that. Leave the denominator alone and concentrate on factoring an h out of the numerator.

Compare yours with:

$$\lim_{h\rightarrow 0}\frac{(x+h)^{2/3}-x^{2/3}}{h}= \lim_{h\rightarrow 0}\left [ \left ( \frac{(x+h)^{2/3}-x^{2/3}}{h}\right ) \left ( \frac{(x+h)^{4/3}+(x+h)^{2/3}x^{2/3}+x^{4/3}}{(x+h)^{4/3}+(x+h)^{2/3}x^{2/3}+x^{4/3}}\right ) \right ]$$

... it does get messy - which is why we don't normally do it that way - you just have to slog through carefully. I find it helps to use a large window and a whiteboard marker.

5. Sep 27, 2013

### Dick

It's not even all that much of a slog, is it? If you remember (a-b)(a^2+ab+b^2)=a^3-b^3, you can immediately replace the numerator with a^3-b^3=(x+h)^2-x^2. That's easy to expand. Then factor out the h and take h->0.

Last edited: Sep 27, 2013
6. Sep 27, 2013

### Simon Bridge

If you are used to this sort of stuff it can be shown in about three steps from here. That would make it just-about "doing the problem for him" - I realize.

If you are just learning - though - there will still be a bunch of false starts, like we've seen already, and it can get confusing keeping track of all the exponents and where the minus signs go.

Making sure students do an exercise like this is supposed to motivate the shortcut approaches. I think MathewsMD has got there already :) The RH parentheses is basically what those a's and b's expand out to and it's the lynchpin. Comparing that with what he got should expose what went wrong.

Building on the power rule for integers, already proved by the definition, and using implicit differentiation, turns this into a bunch of much easier steps... and is more like how math is actually done.

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