Thoughts on this limit ? $\lim_\limits{x\to 3}\dfrac{\frac{1}{3}-\frac{1}{x}}{x-3}$ $$\lim_\limits{x\to 3}\dfrac{\frac{1}{3}-\frac{1}{x}}{x-3}$$ 
Normally, one multiplies by $\frac{3x}{3x}$ to eliminate the complex fraction.  Here, that will not eliminate the problematic denominator term of $(x-3)$.   Neither will subtracting the numerator via LCD.    Or will it?!  Wait a sec...

$$\lim_{x\to3}\frac{\frac{1}{3}-\frac{1}{x}}{x-3}=\lim_{x\to3}\frac{\frac{x}{3x}-\frac{3}{3x}}{x-3}=\lim_{x\to3}\frac{\frac{x-3}{3x}}{x-3}=\lim_{x\to3}\frac{x-3}{3x(x-3)}=\lim_{x\to3}\frac{1}{3x}=\frac{1}{3\cdot3}=\frac{1}{9}$$
