Question

If f is a differentiable function, express the
value of...

Answer 1

\[\lim_{x \rightarrow 1} \frac{ x ^{2}f(x)-f(1) }{ x-1 }\]

Answer 2

in terms of f and f'

Answer 3

use L'Hospitals rule

Answer 4

I disagree with @Zarkon

Use L'Hopital's rule instead!

Answer 5

I did give him a medal though (in a disagreeably disagreeable manner).

Answer 6

lol. OK confession time. I don't what L'Hopital's rule is

Answer 7

@agent0smith could you give me the rule and show me how to approach the problem using it?

Answer 8

Maybe you're not meant to do it that way then...

Answer 9

if you drop the s you should write

L'Hôpital

Answer 10

That's entirely too fancy.

But i can't see how else you'd get an f' in the solution without it.

Answer 11

\[\lim_{x \to1} \frac{ x ^{2}f(x)-f(1) }{ x-1 }\]
\[=\lim_{x \to1} \frac{ x ^{2}[f(x)-f(1)+f(1)]-f(1) }{ x-1 }\]
\[=\lim_{x \to1} \frac{ x ^{2}[f(x)-f(1)]+x^2f(1)-f(1) }{ x-1 }\]
\[=\lim_{x \to1} \frac{ x ^{2}[f(x)-f(1)]+(x^2-1)f(1) }{ x-1 }\]
\[=\lim_{x \to1} \frac{ x ^{2}[f(x)-f(1)]+(x-1)(x+1)f(1) }{ x-1 }\]
\[=\lim_{x \to1} \left[\frac{ x ^{2}[f(x)-f(1)]}{x-1}+\frac{(x-1)(x+1)f(1) }{ x-1 }\right]\]
\[=\lim_{x \to1} \frac{ x ^{2}[f(x)-f(1)]}{x-1}+\lim_{x\to 1}\frac{(x-1)(x+1)f(1) }{ x-1 }\]
\[=1^2\cdot f'(1)+\lim_{x\to 1}(x+1)f(1)\]
\[= f'(1)+2f(1)\]

Answer 12

hey @Zarkon I'm kind of confused how you got at the beginning
\[\lim_{x \rightarrow 1}\frac{ x ^{2}[f(x)-f(1)+f(1)]-f(1) }{ x-1 }\]

Answer 13

\[f(x)=f(x)-f(1)+f(1)\]

Answer 14

the derivative formula is \[\lim_{h \rightarrow 0}\frac{ f(x+h)-f(x) }{ h }\] right? how do you get \[f(x)=f(x)-f(1)+f(1)\] ?

Answer 15

@Zarkon

Answer 16

\[-f(1)+f(1)=0\]

Answer 17

\[f(x)=f(x)+0=f(x)-f(1)+f(1)\]

Answer 18

oh i seeee! that clears things up. thank you!

Answer 19

hey @agent0smith do you think you can help me figure out how @Zarkon got \[\lim_{x \rightarrow 1}\frac{ x ^{2}[f(x)-f(1)]+(x ^{2}-1)f(1) }{ x-1 }\]

Answer 20

where did he get the \[x ^{2}-1 \] ?

Answer 21

He factored it out from the line before

Answer 22

@agent0smith how did he factor it if it's \[\frac{ x ^{2}f(1)-f(1) }{ x-1 }\] ?

Answer 23

Because f(1) is still 1*f(1)

\[\Large x^2f(1) - 1f(1) = (x^2 -1 ) f(1)\]

Answer 24

ooooooh i see! thank you!