# Can You Solve This Limit Without Using L'Hopital's Rule?

• dontdisturbmycircles
In summary: =\frac{\frac{\sqrt[3]{0}+x}{x}}{\frac{\sqrt{0}-x}{x}}=\frac{\frac{\sqrt[3]{0}+x}{x-1}}{\frac{\sqrt{0}-x}{x-1}}=\frac{\frac{\sqrt[3]{0}+x}{x-2}}{\frac{\sqrt{0}-x}{x-2}}=\frac{\frac{\sqrt[3]{0}+x}{x-3}}{\frac{\sqrt{0}-x}{x-3}}=\frac{\frac{\sqrt[3
dontdisturbmycircles
$$\lim_{x->1}\frac{\sqrt[3]{x}-1}{\sqrt{x}-1}$$

I can do it with L'Hospitals, but my teacher said there is also a way to do it without. I tried all sorts of techniques such as multiplying by the conjugate of both the denom/numerator as well as taking $$\sqrt{x}$$ out of both the denom and numerator. I can't see what method my teacher sees that I don't..

I get 2/3 with L'Hospitals by the way.

factor
(a change of variables may help)

Ahhh okay I see, thanks.

Can someone demonstrate how to solve this with a change of variables? I am usually fairly comfortable using a change of variables but I don't see how to apply it here.

For example I can solve lim_{x->inf}e^(-2x) or w/e by making t=-2x. But I don't see what to do in with the problem in the thread.

I can solve it by dividing, but don't see how to use a change of variables.

Last edited:
If you can finish the problem by simply dividing, then you don't need the change of variables. My suggestion to change the variables may help you see how to write the numerator and the denominator so that each can be factored and have some common factors cancel, leaving you with something that you can easily take the limit of.

So, can you see how the numerator and denominator can be separately factored? (Think of polynomials.)

I don't see how to factor them :(. I can rewrite it as t=x-1 so it becomes

$$\lim_{t->0}\frac{t^{1/3}}{t^{1/2}}$$ but that screws it up.

Of course I could simplify to $$t^{-1/6}$$ but I screwed something up somewhere.

Last edited:
Let $$y=\sqrt[6]{x}$$.

$$\lim_{t->1}\frac{t^{2}-1}{t^{3}-1}$$
$$=\lim_{t->1}\frac{(t-1)(t+1)}{(t-1)(t^2+t+1)}$$
$$=\lim_{t->1}\frac{t+1}{t^{2}+t+1}$$
$$=\frac{2}{3}$$

Thanks a lot :) I am still trying to understand your logic for using $$\sqrt[6]{x}$$ though.

I guess it makes sense that given x^n/x^m, replacing one with t^(nm) will make it easier to factor. I still have to think about it a bit.

$$\sqrt[6]{x} = t$$

$$t^{2} = \sqrt[6]{x}^{2} = x^{\frac{1}{3}}$$

$$t^{3} = \sqrt{x}$$

I wanted to get the first line in your previous post... because I knew that I could factor out a common (t-1) factor from a term like (t^k-1). So, I wanted to choose of convenient change of variables get the numerator and denominator to be in the form (t^k-1).

Alongside the above, I wanted to eliminate all of the fractional powers of x.

Ah okay that makes perfect sense. Thanks alot!

L'Hospital's method can be derived from epsilon-delta definitions. Consider:

Assuming
$$x \neq 0$$
We have:
$$\frac{\sqrt[3]{x}-1}{\sqrt{x}-1}=\lim_{x \rightarrow 0} \frac{\sqrt[3]{0+x}-1}{\sqrt{0+x}-1}=\frac{\frac{\sqrt[3]{0+x}-1}{x}}{\frac{\sqrt{0+x}-1}{x}}$$

## 1. What is L'Hopital's rule?

L'Hopital's rule is a mathematical technique used to evaluate limits involving indeterminate forms, such as 0/0 or ∞/∞. It states that if the limit of the quotient of two functions is an indeterminate form, then taking the derivative of both functions and evaluating the limit again will give the same result.

## 2. When should I use L'Hopital's rule?

L'Hopital's rule should only be used when the limit of a quotient of two functions is an indeterminate form, meaning that both the numerator and denominator approach 0 or infinity. It should not be used for limits that can be evaluated by direct substitution.

## 3. How do I know if I can use L'Hopital's rule?

You can use L'Hopital's rule if the limit involves an indeterminate form and both the numerator and denominator are differentiable at the point of evaluation. If either the numerator or denominator is not differentiable, L'Hopital's rule cannot be applied.

## 4. Are there any limitations to using L'Hopital's rule?

Yes, there are limitations to using L'Hopital's rule. It should only be used for limits involving indeterminate forms and cannot be used for other types of limits. Additionally, it can only be used for limits at a single point and cannot be used for limits involving infinity or limits at multiple points.

## 5. Is there an alternative to using L'Hopital's rule?

Yes, there are alternative methods for evaluating limits, such as using algebraic manipulation, trigonometric identities, or series expansions. However, L'Hopital's rule is often the most efficient and straightforward method for evaluating limits involving indeterminate forms.

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