# How to evaluate a limit with roots

• MathewsMD
In summary, the problem is trying to factor out h from the numerator, and using a change of variable and the sum of cubes formula did not help. Next, multiplying the numerator and the denominator by {x+h)^(1/3) + x^(1/3)} to cancel out the roots did not work. Trying variations of the two methods described did not work either. Next, trying to show the derivative by the definition or just by showing the derivative did not work. However, if the former is attempted, then h is found to be in the denominator and the problem becomes solving for x and y. Lastly, using the power rule and the implicit differentiation of integers, it becomes

## Homework Statement

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 into 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!

MathewsMD said:

## Homework Statement

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 into 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!

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).

Dick said:
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).

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.

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

1 person
Simon Bridge said:
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.

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:
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.

## 1. What is a limit with roots?

A limit with roots refers to the process of finding the value that a function approaches as its input variable approaches a certain value, when the function contains a square root or other root expressions.

## 2. How do I evaluate a limit with roots?

To evaluate a limit with roots, you can try to simplify the expression by factoring or using algebraic manipulation. Then, substitute the approaching value into the simplified expression to find the limit.

## 3. What are some common techniques for evaluating limits with roots?

Some common techniques for evaluating limits with roots include using algebraic manipulation, L'Hospital's rule, and rationalizing the expression by multiplying by the conjugate.

## 4. What types of functions typically involve limits with roots?

Functions that involve square roots, cube roots, or other root expressions such as nth roots often require the use of limits with roots to evaluate their behavior as the input variable approaches a certain value.

## 5. Are there any specific tips for evaluating limits with roots?

One tip for evaluating limits with roots is to always check for the existence of a removable discontinuity. If there is a common factor between the numerator and denominator that can be canceled out, the limit can be evaluated using direct substitution.