Question

If f'(1) = 2 , f(1) = 4
then limit 3h/ f(1+h) -2
h - > 0

Answer 1

\[\lim_{h \rightarrow 0}(\frac{3h}{f(1+h)}-2) \text{ or } \lim_{h \rightarrow 0}(\frac{3h}{f(1+h)-2}) ?\]
which of these is what you mean?

Answer 2

direct sub can be done on either one of these
we won't even need f'(1)=2

Answer 3

the second

Answer 4

the f'(1)=2 tells us f() is differentiable at x=1, and that allows us to assume
the limit f(1+h) --> f(1) as h-->0

Answer 5

limit 3h/ f(1+h) -2 = ..... <- forgot the equal

Answer 6

I don't know F(x) :/

Answer 7

have you tried that direct subsitution thingy yet?

Answer 8

Show me what do you mean

Answer 9

like and yeah f'(1)=2 does actually say f is diff at x=1
which means it is continuous at x=1
so yeah phi was correct to say that is why we can assume as h->0
f(1+h)->f(1)=2

Answer 10

just replace the yh's with 0

Answer 11

h's*

Answer 12

The book solution is 3/4

Answer 13

not possible

Answer 14

Possible :D

Answer 15

unless you have gave the wrong problem

Answer 16

If f was a function : F'(1) =2 , F(1) = 4

so : \[\lim_{h \rightarrow 0 } \frac{ 3h }{ F(1+h) -2 }\] = ........

Answer 17

Solution : [3/4]

Answer 18

oh.

I guess we have to do this
\[ f'(1) = \lim_{h\rightarrow 0} \frac{ f(1+h) - f(1)}{h} \]

Answer 19

That's what I tried I got 3h/ 2 + 2h

Answer 20

What I was suspicious of is that maybe f(1) = 2 and F'(1) =4 ?
so 3h/ 4h = 3/4 but I am always wrong.

Answer 21

yes, I think we have to assume f'(1)= 4 and f(1)=2, so we can do this
\[ \lim_{h \rightarrow 0}\left( \frac{3h}{f(1+h)-2}\right) \\
= 3 \lim_{h \rightarrow 0} \frac{1}{ \frac{f(1+h)-2}{h} } \\
= 3 \lim_{h \rightarrow 0} \frac{1}{ \frac{f(1+h)-f(1)}{h} } \
\]

Answer 22

that turns into
3 * 1/f'(1)
so if f'(1) = 4 we get 3/4

Answer 23

I thought f'(1)=2 ?

Answer 24

and f(1)=4

Answer 25

It is.

Answer 26

\[\lim_{h \rightarrow 0}(\frac{3h}{f(1+h)-2}) \]
wait
so if it was the other way around
f(1)=2 and f'(1)=4
if we tried direct sub we would get 0/0
then we would apply l'hospital
\[\lim_{h \rightarrow 0}\frac{3}{f'(1+h)} \\ =\frac{3}{4}\]

Answer 27

but the other way around it is 0

Answer 28

yes, it looks like they switched the f and f' values.
otherwise I don't see how to get the book answer.

Answer 29

ditto!

Answer 30

So the book substituted them ?

Answer 31

you can use the definition of the derivative (see above)
or use L'hopital (see above)

Answer 32

Thanks both of you guys.