What is the values of $a$ and $b$ without using the L'Hôpital's Rule Suppose that $$\frac{(2x)^x-2} {a(x-1)+b(x-1)^2}\to 1$$ as $x \to 1$. ThenWhat is the values of $a$ and $b$ without using the L'Hôpital's Rule?
Thanks for your help!

Factor an $x-1$ out of the denominator to get $$\frac{(2x)^x-2} {a(x-1)+b(x-1)^2} = \frac{(2x)^x -  2}{x-1} \cdot \frac{1}{a + b(x-1)}$$ so that (as long as $a \not= 0$)$$\lim_{x \to 1} \frac{(2x)^x-2} {a(x-1)+b(x-1)^2} = \left.\frac{d}{dx} (2x)^x \right|_{x=1} \cdot \frac 1a.$$
But $\dfrac{d}{dx} (2x)^x = (2x)^x (1 + \log 2x)$, and at $x=1$ equals $2(1 + \log 2)$. Thus $$\lim_{x \to 1} \frac{(2x)^x-2} {a(x-1)+b(x-1)^2} = \frac{2(1 + \log 2)}{a}.$$
