Solving $\lim_{x\rightarrow0+} x^2 \ln x$ with L'Hoptial's Rule

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In summary, the conversation was about finding the limit of x^2 ln x as x approaches 0 using L'Hopital's rule. There was some confusion about the derivatives of f(g) and f(h), but it was clarified that f(x) = ln x and g(x) = x^-2, and the derivatives were found using the quotient rule. The answer to the problem was determined to be -2.
  • #1
DeadWolfe
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I was asked to find

[tex]\lim_{x\rightarrow0+} x^2 \ln x [/tex]

Using L'Hoptial's rule.

I guess that I have to divide by one of these things to the power of negative one, and then take the derivative of top and bottom, but this results in either dividing by zero if I put x^2 on the bottom, or not losing ln x if I put it on the bottom, because of the chain rule.

I really am doubting that L'Hopital is the way to go about this, but that's what I was told on a test today.

Any ideas?
 
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  • #2
You aren't dividing by 0 by putting x^2 on the bottom. Remember, this is a limit, the values aren't fixed.

[tex] \frac{ln(x)}{1/x^2} [/tex]

[tex] \frac{1/x}{-2/x^3} [/tex]

[tex] -x^2/2 [/tex]

[tex] 0 [/tex]
 
  • #3
DeadWolfe said:
I was asked to find

[tex]\lim_{x\rightarrow0+} x^2 \ln x [/tex]

Using L'Hoptial's rule.

I guess that I have to divide by one of these things to the power of negative one, and then take the derivative of top and bottom, but this results in either dividing by zero if I put x^2 on the bottom, or not losing ln x if I put it on the bottom, because of the chain rule.

I really am doubting that L'Hopital is the way to go about this, but that's what I was told on a test today.

Any ideas?

[tex]\lim_{x\rightarrow0+} \frac{x^3 \ln x}{x} [/tex]
 
  • #4
StatusX - thank you, I tihnk that works

Zlex - do you see how that method will not eliminate the ln x because of the product rule?
 
  • #5
Hey everyone. I know this question is old but I'm trying to solve exactly the same problem. Could anyone explain it to me a little better? I'm having trouble seeing how StatusX got the answer.

I understand L'Hopital's rule but how the derivatives of f(g) and f(h) were found is confusing me.
 
Last edited:
  • #6
The derivatives of f(g) and f(h)? Do you mean f(x) and g(x)? Either way they did the following:

[tex]x^2 \ln x=\frac{\ln x}{x^{-2}} \Rightarrow f(x)=\ln x, g(x)=x^{-2} \Rightarrow \lim_{x \to 0+}x^2 \ln x=\lim_{x \to 0+} \frac{f(x)}{g(x)}=\lim_{x \to 0+} \frac{f'(x)}{g'(x)}=\lim_{x \to 0+} \frac{\frac{1}{x}}{-2 x^{-3}}=\lim_{x \to 0+} -2 x^2[/tex].
 
  • #7
The numerator, in Statusx's form, is ln(x), and the denominator is 1/x2= x-2.

The derivative of ln(x) is a "standard form", derived from the fact that the derivative of ex is ex. d(ln(x))/dx= 1/x.

The derivative of x-2 is, of course, from the power rule: (-2)x-2-1= -2x-3= -2/x3.
 
  • #8
Thanks guys. That helps a lot. I still don't know every little trick of derivatives and such.
 

1. What is L'Hopital's Rule?

L'Hopital's Rule is a mathematical principle that states that the limit of a ratio of two functions is equivalent to the limit of the ratio of their derivatives, provided that both functions approach the same value as x approaches a certain point.

2. How is L'Hopital's Rule applied to solve limits?

To apply L'Hopital's Rule, you must first take the derivative of both the numerator and denominator of the given function. Then, you can evaluate the limit of the new function, which will often be easier to solve. If the new limit still results in an indeterminate form, you can repeat the process until a definitive answer is reached.

3. Can L'Hopital's Rule be used for all limits?

No, L'Hopital's Rule can only be applied to limits that result in indeterminate forms, such as 0/0 or ∞/∞. If the limit does not result in an indeterminate form, L'Hopital's Rule cannot be used.

4. What is the purpose of using L'Hopital's Rule?

The main purpose of L'Hopital's Rule is to simplify the evaluation of limits that would otherwise be difficult or impossible to solve. It allows us to use the properties of derivatives to find the limit of a function in cases where direct substitution would not work.

5. How is L'Hopital's Rule used to solve the limit of x^2 ln x as x approaches zero from the positive side?

To solve this limit, we first take the derivative of both x^2 and ln x, which results in 2x and 1/x, respectively. Then, we can rewrite the given limit as the limit of (2x)/1 as x approaches zero. This limit is easily solved by direct substitution, resulting in a final answer of 0.

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