Homework Help: Limit problem

1. Dec 8, 2018

Krushnaraj Pandya

1. The problem statement, all variables and given/known data
$\lim x-a \frac {{a^x-{x^a}}}{{x^x}-{a^a}} = -1$ then a is?

2. Relevant equations
L'Hospital rule

3. 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?

2. Dec 8, 2018

Math_QED

Limit of $x \to a$ I assume?

3. Dec 8, 2018

yes

4. Dec 8, 2018

Math_QED

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

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

5. Dec 8, 2018

Krushnaraj Pandya

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. Dec 8, 2018

Math_QED

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. Dec 8, 2018

Krushnaraj Pandya

That was very helpful, thank you :D

8. Dec 8, 2018

Ray Vickson

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. Dec 9, 2018

Krushnaraj Pandya

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...)