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$\displaystyle{\lim_{ x \rightarrow a}}\;\frac{a^x-x^a}{x-a}\;\;$

I don't know what i've been doing wrong. I've triying to solve it, using this remarkable limit:

$\displaystyle{\lim_{ x \rightarrow 0}}\; \frac{a^{kx}-1}{x}=k\;ln(a)\;\;$ ln is the natural log

Since the limit has to approach 0, to use the remarkable, i did a variable change.

Let $u=x-a$, if $x\rightarrow a$ then $u \rightarrow 0$, this means that $x=u+a$

So:

$\displaystyle{\lim_{ u \rightarrow 0}}\; \dfrac{a^{u+a}-(u+a)^{a}}{u}$

$=\displaystyle{\lim_{ u \rightarrow 0}}\; \dfrac{a^{u+a}-(u+a)^{a}-1+1}{u}\;\;$ (Substrac and add 1 to numerator)

$=\displaystyle{\lim_{ u \rightarrow 0}}\; \dfrac{(a^{u+a}-1)-((u+a)^{a}-1)}{u}\;\;$ (Common factor (-1))

$=\displaystyle{\lim_{ u \rightarrow 0}}\; \dfrac{a^{u+a}-1}{u}-\dfrac{(u+a)^{a}-1}{u}$

$=\displaystyle{\lim_{ u \rightarrow 0}}\; \dfrac{\dfrac{(u+a)[a^{u+a}-1]}{u+a}}{u}-\dfrac{\dfrac{a[(u+a)^{a}-1]}{a}}{u}\;\; $ (common factor (u+a) and (a))

I don't know if the steps that come next are right

$=\displaystyle{\lim_{ u \rightarrow 0}}\; \dfrac{(u+a)ln(a)}{u}-\dfrac{(a)ln(u+a)}{u}\;\;$ (Apply remarkable limit)

$=\displaystyle{\lim_{ u \rightarrow 0}}\; \dfrac{ln(a^{u+a})}{u}-\dfrac{ln(u+a^{a})}{u}$

And here is where I'm lost, i don't know what to do next. Wolfram alpha says that:

$\displaystyle{\lim_{ x \rightarrow a}}\;\frac{a^x-x^a}{x-a}= a^aln(a)-a^a\;\;$

I can't use l'Hôpital rule

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  • $\begingroup$ You started off correctly, but made the wrong split. You have to add and subtract $a^a$ and not $1$. $\endgroup$ – Paramanand Singh Feb 13 '18 at 3:04
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You can proceed as $$\frac{a^{u+a} - (u+a) ^{a} - a^a+a^a} {u} =\frac{a^{u+a} - a^a} {u} - \frac{(u+a) ^{a} - a^a} {u} $$ The first term can be written as $$a^a\cdot\frac{a^u-1}{u}$$ which tends to $a^a\log a$ via the remarkable limit. The second term needs to be put back in the form of $x$ to get $$\frac{x^a-a^a} {x-a} $$ and this tends to $aa^{a-1}=a^a$ via another remarkable limit $$\lim_{x\to a} \frac{x^n-a^n} {x-a} =na^{n-1}$$ Combining the above we get final limit as $a^a\log a-a^a$.


Apart from making the wrong split using $-1+1$ (this is correct mathematically but does not help to reach the goal) your approach has some mistakes which are mathematically incorrect. Thus you multiply / divide by $(u+a) $ and then incorrectly apply the remarkable limit. The limit applies only when denominator tends to $0$ but here denominator $u+a\to a$. Moreover in this process you are not taking full limit, rather the limit of part of an expression. This is another thing which is forbidden.

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  • $\begingroup$ Thanks for your answer sir, you are very kind. $\endgroup$ – Diego Sánchez Feb 13 '18 at 12:53

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