Solving a Limit Problem using L'Hospital Rule

  • Thread starter Krushnaraj Pandya
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In summary: I have practiced mostly with L'Hospital, although I'm aware of the expansions of...standard functions?
  • #1
Krushnaraj Pandya
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Homework Statement


## \lim x-a \frac {{a^x-{x^a}}}{{x^x}-{a^a}} = -1## then a is?

Homework Equations


L'Hospital rule

The Attempt at a Solution


Using LHR we can write numerator as ##\frac{a^x ln{a}-ax^{a-1}}{x^xln(x+1)}##
plugging x=a and equating to -1 gives 1-ln(a)=ln(a+1); so 1=ln(a(a+1)) but this gives an incorrect answer. Where am I going wrong?
 
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  • #2
Limit of ##x \to a## I assume?
 
  • #3
Math_QED said:
Limit of ##x \to a## I assume?
yes
 
  • #4
Krushnaraj Pandya said:
yes

Maybe this is your mistake:

##(x^x)' = x^x(\ln x + 1)##

Not: ##(x^x)' = x^x \ln(x+1)## like you wrote.
 
  • #5
Math_QED said:
Maybe this is your mistake:

##(x^x)' = x^x(\ln x + 1)##

Not: ##(x^x)' = x^x \ln(x+1)## like you wrote.
I routinely make such mistakes while studying math, I know all the correct things and math could be enjoyable and I could feel like I'm good at it if only I stopped making them...do you have any suggestions on how I can stop making these silly mistakes?
and Thank you very much for your help :D
 
  • #6
Krushnaraj Pandya said:
I routinely make such mistakes while studying math, I know all the correct things and math could be enjoyable and I could feel like I'm good at it if only I stopped making them...do you have any suggestions on how I can stop making these silly mistakes?
and Thank you very much for your help :D

I'm glad I could help.

For your question, check out this wonderful article written by someone on this forum who is more experienced/knowledgeable than me, where he adresses the issue (the thing is written for exam situations, but really applies to any situation).

https://www.physicsforums.com/insights/10-math-tips-save-time-avoid-mistakes/
 
  • #7
Math_QED said:
I'm glad I could help.

For your question, check out this wonderful article written by someone on this forum who is more experienced/knowledgeable than me, where he adresses the issue (the thing is written for exam situations, but really applies to any situation).

https://www.physicsforums.com/insights/10-math-tips-save-time-avoid-mistakes/
That was very helpful, thank you :D
 
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  • #8
Krushnaraj Pandya said:

Homework Statement


## \lim x-a \frac {{a^x-{x^a}}}{{x^x}-{a^a}} = -1## then a is?

Homework Equations


L'Hospital rule

The Attempt at a Solution


Using LHR we can write numerator as ##\frac{a^x ln{a}-ax^{a-1}}{x^xln(x+1)}##
plugging x=a and equating to -1 gives 1-ln(a)=ln(a+1); so 1=ln(a(a+1)) but this gives an incorrect answer. Where am I going wrong?

You can set ##x = a+h## to write your fraction as
$$\text{fraction} = \frac{a^{a+h} - (a+h)^a}{(a+h)^{a+h} - a^a}, $$
then take the limit as ##h \to 0## using l'Hospital's rule. However, my personal preference would be to expand the numerator and denominator in a few powers of small ##|h|##. For example, the numerator is ##N_h = a^a a^h - (a+h)^a.## We can expand the first term by setting ##a = e^{\ln a}##, so that ##a^h = e^{h \ln(a)}##, which is easily expanded in powers of ##h##; keeping only linear terms in ##h## is good enough in this problem. The second term is ##(a+h)^a,## which has a binomial expansion in ##h##. Again, stopping at terms linear in ##h## is good enough in this problem. The denominator involves ##(a+h)^{a+h},## which can be written as ##e^{(a+h) \ln(a+h)} ##. You can expand the exponent ##r = (a+h) \ln(a+h)## in powers of ##h## (again, terms linear in ##h## are good enough); then you can expand the exponential ##e^r## as ##1 + r + \cdots##, where ##\cdots## stands for higher powers in ##r##. Now substitute your expansion of ##r## in terms of ##h##.

Of course, in principle, this "expansion" method is really just l'Hospital's rule in disguise, but I find it faster and easier than direct application of l'Hospital. (However, it all depends on how familiar you are with the expansions of the standard functions.)
 
  • #9
Ray Vickson said:
You can set ##x = a+h## to write your fraction as
$$\text{fraction} = \frac{a^{a+h} - (a+h)^a}{(a+h)^{a+h} - a^a}, $$
then take the limit as ##h \to 0## using l'Hospital's rule. However, my personal preference would be to expand the numerator and denominator in a few powers of small ##|h|##. For example, the numerator is ##N_h = a^a a^h - (a+h)^a.## We can expand the first term by setting ##a = e^{\ln a}##, so that ##a^h = e^{h \ln(a)}##, which is easily expanded in powers of ##h##; keeping only linear terms in ##h## is good enough in this problem. The second term is ##(a+h)^a,## which has a binomial expansion in ##h##. Again, stopping at terms linear in ##h## is good enough in this problem. The denominator involves ##(a+h)^{a+h},## which can be written as ##e^{(a+h) \ln(a+h)} ##. You can expand the exponent ##r = (a+h) \ln(a+h)## in powers of ##h## (again, terms linear in ##h## are good enough); then you can expand the exponential ##e^r## as ##1 + r + \cdots##, where ##\cdots## stands for higher powers in ##r##. Now substitute your expansion of ##r## in terms of ##h##.

Of course, in principle, this "expansion" method is really just l'Hospital's rule in disguise, but I find it faster and easier than direct application of l'Hospital. (However, it all depends on how familiar you are with the expansions of the standard functions.)
I have practiced mostly with L'Hospital, although I'm aware of the expansions of standard functions. (The trouble is I can't remember and retain them...)
 

Related to Solving a Limit Problem using L'Hospital Rule

1. What is the L'Hospital Rule?

The L'Hospital Rule is a method used to evaluate limits of indeterminate forms in calculus. It states that if the limit of a function f(x) divided by g(x) is an indeterminate form (such as 0/0 or ∞/∞), then the limit of f(x)/g(x) is equal to the limit of the derivatives of f(x) and g(x) as x approaches the same value.

2. When should I use the L'Hospital Rule?

The L'Hospital Rule should be used when evaluating a limit that results in an indeterminate form. This typically occurs when the limit of the numerator and denominator both approach 0 or infinity. For example, it can be used to evaluate limits involving rational functions, exponential functions, and trigonometric functions.

3. Are there any restrictions when using the L'Hospital Rule?

Yes, there are a few restrictions when using the L'Hospital Rule. First, the function must be continuous and differentiable in a neighborhood around the limit point. Second, the limit must be in the form of an indeterminate form. Third, the limit must be approached by real numbers. Lastly, the limit must be taken as x approaches a real number or infinity.

4. How do I apply the L'Hospital Rule?

To apply the L'Hospital Rule, take the derivative of both the numerator and denominator of the original function. Then, evaluate the limit of the derivatives as x approaches the same value as the original limit. If the resulting limit is still an indeterminate form, continue to take derivatives until the limit can be evaluated. Lastly, if the limit of the derivatives does not exist, then the original limit does not exist.

5. Can the L'Hospital Rule be used for one-sided limits?

Yes, the L'Hospital Rule can be used for one-sided limits. The only difference is that the derivatives must be taken from the same side as the original limit. For example, if the original limit is approaching from the left, then the derivatives must also be taken from the left. This is important because the derivatives may be different depending on the direction of approach.

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