Evaluating $\lim_{y\rightarrow x}\frac{9^{x+y}-9^{2y}}{3^{x+y}-3^{2x}}$

Question

I have the following limit to solve, and without resorting to l'Hôpital:

$$\lim_{y\rightarrow x}\frac{9^{x+y}-9^{2y}}{3^{x+y}-3^{2x}}$$

What I tried to do:

$$\lim_{y\rightarrow x}\frac{9^{x+y}-9^{2y}}{3^{x+y}-3^{2x}}=\frac{9^{2x}-9^{2x}}{3^{2x}-3^{2x}}=\frac{(3^{2x}-3^{2x})(3^{2x}+3^{2x})}{3^{2x}-3^{2x}}=3^{2x}+3^{2x}=2\cdot 9^x$$

But the answer seems to be $-2\cdot9^x$. Where is my error and how should I approach the limit? Thanks in advance.

Answer 1

$$\lim_{y\rightarrow x}\frac{9^{x+y}-9^{2y}}{3^{x+y}-3^{2x}}=\\ \lim_{y\rightarrow x}\frac{9^y.(9^{x}-9^{y})}{3^x.(3^{y}-3^{x})}=\\ \lim_{y\rightarrow x}\frac{9^y.(3^{x}-3^{y})(3^{x}+3^{y})}{3^x.(3^{y}-3^{x})}=\\ \lim_{y\rightarrow x}\frac{-9^y.(3^{y}-3^{x})(3^{x}+3^{y})}{3^x.(3^{y}-3^{x})}=\\ \lim_{y\rightarrow x}\frac{-9^y.(3^{x}+3^{y})}{3^x}=\\ \frac{-9^x.(3^{x}+3^{x})}{3^x}=\\\frac{-9^x.(2.3^{x})}{3^x}=\\-2 \cdot 9^x$$

Answer 2

Hint. Let $u=3^{y-x}$ then $$\frac{9^{x+y}-9^{2y}}{3^{x+y}-3^{2x}}=\frac{9^{2x}(u^2-u^4)}{9^{x}(u-1)}=-9^xu^2(1+u).$$ Now note that $u\to 1$.