Solving $\displaystyle \lim_{x \rightarrow 1} \frac{ab^x+ba^x}{x-1}$

[Sleek]

Homework Statement

$$\displaystyle \lim_{x \rightarrow 1} \frac{ab^x+ba^x}{x-1}$$

Homework Equations

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The Attempt at a Solution

My attempt was to let x-1 = u

As x -> 1, x-1 -> 0, therefore as x-> 1, u -> 0.

So I obtained something which looked simple.

$$\displaystyle ab \lim_{u \rightarrow 0} \frac{a^u + b^u}{u}$$

Had it been a^u-b^u, Adding and subtracting 1 and splitting the expression into (a^u-1)/u and (b^u-1)/u would help solve the limit, which is loga+logb. Thus the answer becomes ab log(ab).

But since there is a plus sign, I'm confused on what I could do further. Incidentally, th answer to the question is ab log(b/a), which is quite close to the previous one, and does imply a sign change between the log expressions.

Can anyone verify my substitution and just give a subtle hint towards the direction I have to look? Also, is there a place where I can find proofs of standard forms like lim(x -> 0) (a^x-1)/x = ln a etc.?

Thanks,
Sleek.

 

[AKG]

The numerator goes to 2ab. If either a or b is 0, the numerator is identically 0 (for positive x, in particular for x near 1) so the limit is 0. Otherwise, 2ab is finite non-zero whereas the denominator goes to positive infinity as x goes to 1 from the right and negative infinity from the other direction, meaning the limit doesn't exist. Perhaps you wrote the problem out wrong?