Question

Use the alternate limit definition,
Limit as x approaches a f(x) - f(a) /x-a
With a = 2, to find the value of f ′(2).

Answer 1

What is the Alternate Limit Definition

Answer 2

I'd recommend that you google terms such as "alternate limit definition" on your own, because the look-up skills you'd develop by doing so will be extremely important as your education progresses.

Here are some of the returns I got from googling "alternative limit definition."

https://www.google.com/search?q=lternate+Limit+Definition&oq=lternate+Limit+Definition&aqs=chrome..69i57j0l4.383j0j7&sourceid=chrome&ie=UTF-8

Answer 3

That has not helped me

Answer 4

I feel like there is something lacking in your question like what is f(x)=?

Anyways here an example:
Say
\[f(x)=\frac{1}{x+3} \\ f'(2)=\lim_{x \rightarrow 2} \frac{f(x)-f(2)}{x-2} \text{ I replaced } a \text{ with } 2 \\ f'(2)=\lim_{x \rightarrow 2} \frac{\frac{1}{x+3}-\frac{1}{2+3}}{x-2} \text{ I evaluate } f(x) \text{ and } f(2) \text{ using } f(x)=\frac{1}{x+3} \\ f'(2)=\lim_{x \rightarrow 2} \frac{\frac{1}{x+3}-\frac{1}{5}}{x-2} \text{ now let's get rid of the compound fraction } \]
\[
\text{ Multiply numerator and denominator by } 5(x+3) \]

\[f'(2)=\lim_{x \rightarrow 2} \frac{\frac{1}{x+3}-\frac{1}{5}}{x-2} \cdot \frac{5 (x+3)}{5(x+3)} \\ f'(2)=\lim_{x \rightarrow 2} \frac{\frac{1}{x+3} \cdot 5(x+3)-\frac{1}{5} \cdot 5 (x+3)}{(x -2) \cdot 5(x+3)} \text{ I distributed on top}\]


\[f'(2)=\lim_{x \rightarrow 2} \frac{5-(x+3)}{(x-2) \cdot 5(x+3)}\]

\[f'(2)=\lim_{x \rightarrow 2} \frac{-x+2}{(x-2) \cdot 5(x+3)} \\ f'(2)=\lim_{x \rightarrow 2} \frac{-(x-2)}{(x-2) \cdot 5(x+3)} \\ \text{ notice we can finally cancel the thing that would make the bottom 0}\]

\[f'(2)=\lim_{x \rightarrow 2} \frac{-1}{5 (x+3)} \\ \text{ We can finally plug in } 2 \\ f'(2)=\frac{-1}{5(2+3)}=\frac{-1}{5(5)}=\frac{-1}{25}\]

Answer 5

THANK YOU!!!