Evaluating a limit using L'Hopital

In summary, the problem hints suggest that I use L'Hopital to evaluate the limit of the denominator, but taking the derivative will be a huge pain.
  • #1
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Homework Statement


For[tex]Q(x) = x^k + \sum_{n=0}^{k-1} a_n x^n[/tex]

Find [tex]\lim_{x→∞}\left({Q(x)^{\frac{1}{k}} - x}\right)[/tex]


Homework Equations


L'Hopital


The Attempt at a Solution


[tex]Q(x) - x^k = \sum_{n=0}^{k-1} a_n x^n[/tex]
[tex]Q(x) - x^k = (Q(x)^{\frac{1}{k}})^k - x^k = \left(Q(x)^{\frac{1}{k}} - x\right)\left([Q(x)^{\frac{1}{k}}]^{k-1} + x[Q(x)^{\frac{1}{k}}]^{k-2} + ... + x^{k-2}Q(x)^{\frac{1}{k}} + x^{k-1} \right)[/tex]

[tex]Q(x)^{\frac{1}{k}} - x = \frac{Q(x) - x^k}{[Q(x)^{\frac{1}{k}}]^{k-1} + x[Q(x)^{\frac{1}{k}}]^{k-2} + ... + x^{k-2}Q(x)^{\frac{1}{k}} + x^{k-1}}[/tex]

Then [itex]\lim_{x→∞}{\left(Q(x)^{\frac{1}{k}} - x\right)} = \lim_{x→∞}\frac{f(x)}{g(x)}[/itex], and if I can show [itex]\lim_{x→∞}g(x) = ∞[/itex], I can use L'Hopital.

Am I headed in the right direction? I can see this either working out nicely after a lot of work or being an incredible waste of time. Am I overlooking a simpler way to evaluate this limit?
 
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  • #2
Try to factor things as follows:

[tex]x^k-Q(x)=x^{k-1}\left(a_{k-1}+\frac{a_{k-2}}{x}+...+\frac{a_0}{x^{k-1}}\right)[/tex]

and

[tex](Q(x)^\frac{1}{k})^{k-1}=x^{k-1}(\sqrt[k]{1+\frac{a_{k-1}}{x}+\frac{a_{k-2}}{x^2}+...+\frac{a_0}{x^{k}}})^{k-1}[/tex]

Try to factor all terms like that and use that things like [itex]\frac{1}{x^p}\rightarrow 0[/itex].
 
  • #3
Okay, so that will help me show that the limit of the denominator is ∞, but then taking the derivative is going to be a huge pain...
 
  • #4
No, you don't need to differentiate anything!

Do something like this

\begin{align}
\lim_{x\rightarrow +\infty}{\frac{x^3+x}{(\sqrt{x^2+x+1})^3}}
&= \lim_{x\rightarrow +\infty} {\frac{ x^3(1+\frac{1}{x^2}) }{ x^3(\sqrt{1+\frac{1}{x}+\frac{1}{x^2}})^3} }\\
&= \lim_{x\rightarrow +\infty} {\frac{ 1+\frac{1}{x^2} }{(\sqrt{1+\frac{1}{x}+\frac{1}{x^2}})^3} }\\
&= 1
\end{align}
 
  • #5
Oh, okay. The problem hint suggested I use L'Hopital, so I guess I was thinking "in the box" when it came to evaluating the limit.

Thanks for the suggestions!
 

What is L'Hopital's rule and when is it used?

L'Hopital's rule is a mathematical tool used to evaluate the limit of a function that approaches an indeterminate form, such as 0/0 or ∞/∞. It can be used when applying direct substitution to the limit results in an undefined answer.

What are the steps to evaluate a limit using L'Hopital's rule?

The steps to evaluate a limit using L'Hopital's rule are:

  1. Identify the indeterminate form of the limit.
  2. Take the derivative of both the numerator and denominator of the function.
  3. Evaluate the limit again using the new derivatives.

Can L'Hopital's rule be used for all limits?

No, L'Hopital's rule can only be used for limits that approach an indeterminate form, such as 0/0 or ∞/∞. It cannot be used for limits that approach a finite value or ∞ without an indeterminate form.

What are some common mistakes when using L'Hopital's rule?

Some common mistakes when using L'Hopital's rule include:

  • Applying the rule to limits that do not approach an indeterminate form.
  • Not taking the derivative of both the numerator and denominator of the function.
  • Using L'Hopital's rule multiple times without checking if the limit is still indeterminate.

Are there any alternatives to using L'Hopital's rule for evaluating limits?

Yes, there are alternative methods for evaluating limits, such as factoring, simplifying, and using trigonometric identities. However, L'Hopital's rule is often the most efficient method for evaluating limits that approach an indeterminate form.

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