Solving a Limit Problem using L'Hospital Rule

[Krushnaraj Pandya]

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?

 

[member 587159]

Limit of $x \to a$ I assume?

 

[Krushnaraj Pandya]

Limit of $x \to a$ I assume?

yes

 

[member 587159]

yes

Maybe this is your mistake:

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

Not: $(x^x)' = x^x \ln(x+1)$ like you wrote.

 

[Krushnaraj Pandya]

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

 

[member 587159]

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/

 

[Krushnaraj Pandya]

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

 

[Ray Vickson]

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.)

 

[Krushnaraj Pandya]

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...)