What is the limit of the following problem?

Question

I need to find the limit of the following function : $$\lim_{x\rightarrow 0} \ \frac{a^{\ x} \ b^{ \ x} - b^{ \ x} - a^{\ x} +1}{x^{ \ 2}}$$

I know that it is in $\frac{0}{0}$ form so I have applied L'Hospital rule as seen below:

$$\lim_{x\rightarrow 0}\frac{(ab)^{x} \ log(b) + (ab)^{x} \ log(a) - b^x \ log(b) - a^x \ log(a)}{2x} $$

Applying the L'Hospital rule again I ended up with: $$\lim_{x\rightarrow 0}\frac{(ab)^{x}\log(ab) \left [ log(a) + log(b) \right ]\ - \ b^{x} \ (log \ b)^{2} \ - a^{x}\ (log \ a)^{2}}{2} \ $$

$$ =\frac{(log \ a)^{2} + (log \ b)^{2} + 2log(a)log(b) \ - (log \ a)^{2} - (log \ b)^{2}}{2}\ $$ $$=log(a)log(b)$$

Is the answer and method correct? Also if correct, is there a shorter way of solving it?

Answer 1

The method is correct, but you can observe that $$ a^xb^x-a^x-b^x+1=(a^x-1)(b^x-1) $$ and so you just need to compute $$ \lim_{x\to0}\frac{a^x-1}{x} $$ which is the derivative at zero of $f(x)=a^x$, which is well known to be $\log a$. Thus $$ \lim_{x\to0}\frac{a^x-1}{x}\frac{b^x-1}{x}=(\log a)(\log b) $$

Answer 2

Yes it's correct.

I think, it's better to make the following: $$\frac{a^xb^x-a^x-b^x+1}{x^2}=\frac{a^x-1}{x}\cdot\frac{b^x-1}{x}\rightarrow\ln{a}\ln{b}.$$

Answer 3

Yes it is correct, more simply by standard limit we have

$$\frac{a^{\ x} \ b^{ \ x} - b^{ \ x} - a^{\ x} +1}{x^{ \ 2}}= \frac{b^x(a^{\ x} \ -1) - (a^{\ x} -1)}{x^{ \ 2}}=\frac{a^{\ x} \ -1}{x}\frac{b^{\ x} \ -1}{x}\to \log a \log b$$

Answer 4

$$\frac{a^xb^x-a^x-b^x+1}{x^2}=\frac{a^x-1}{x}\cdot\frac{b^x-1}{x}$$ and $$\lim_{x\to 0}\frac{a^x-1}{x}=\frac{d}{dx}(a^x)\bigg|_{x=0}$$