Q. 6 - Calculus

Question

Explain why the limits $\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$ and $\underset{z\to x}{\lim }\frac{f\left(z\right)-f\left(x\right)}{z-x}$ are each initially in the form $\frac{0}{0}$. Why would cancelling a common factor of $h$ or $z-x$ be likely to resolve this indeterminate form?

Step-by-Step Solution

Answer

The value of $\frac{0}{0}$ can't be defined or it can be said that its value doesn't exist. Therefore to define a limit at that particular point the factors should be cancelled from both numerator and denominators.

Step 1. Given information

We have to say why these $\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$ and $\underset{z\to x}{\lim }\frac{f\left(z\right)-f\left(x\right)}{z-x}$ are each initially in the form $\frac{0}{0}$ and why would cancelling a common factor of $h$ or $z-x$ be likely to resolve this indeterminate form.

Step 2. Reason for the given question

The value of $\frac{0}{0}$ can't be defined or it can be said that its value doesn't exist. Therefore to define a limit at that particular point the factors should be cancelled from both numerator and denominators.