Answer: [tex]-\frac{9}{x^2}[/tex]
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Work Shown:
[tex]\displaystyle \lim_{\Delta x \to 0^{+}} \frac{\frac{9}{x+\Delta x} - \frac{9}{x}}{\Delta x}\\\\\\\displaystyle \lim_{\Delta x \to 0^{+}} \frac{\frac{9x}{x(x+\Delta x)} - \frac{9(x+\Delta x)}{x(x+\Delta x)}}{\Delta x}\\\\\\\displaystyle \lim_{\Delta x \to 0^{+}} \frac{\frac{9x-9(x+\Delta x)}{x(x+\Delta x)}}{\Delta x}\\\\\\\displaystyle \lim_{\Delta x \to 0^{+}} \frac{9x-9x-9\Delta x}{x\Delta x(x+\Delta x)}\\\\\\[/tex]
[tex]\displaystyle \lim_{\Delta x \to 0^{+}} \frac{-9\Delta x}{x\Delta x(x+\Delta x)}\\\\\\\displaystyle \lim_{\Delta x \to 0^{+}} \frac{-9}{x(x+\Delta x)}\\\\\\\displaystyle \frac{-9}{x(x+0)}\\\\\\\displaystyle -\frac{9}{x^2}\\\\\\[/tex]
The trick here is to first simplify the expression so that the [tex]\Delta x[/tex] term in the very lower denominator cancels out. This is to avoid dividing by zero. Once this division by zero error is avoided, we can replace the delta x term with 0 and evaluate the limit as [tex]\Delta x[/tex] approaches 0 from the right or positive direction.
