Question

Find value of c, and compute g'(1)

Answer 1

for the function to be continuous at x = 1, you must have :

\(\large \lim \limits_{x \to 1 } f(x) = f(1)\)

Answer 2

\(\large \lim \limits_{x \to 1}~ \frac{\ln x}{x-1} = c\)

find c

Answer 3

if I do that c=0, which is not the case, since c=1, how do i get c=1?

Answer 4

how do u get c = 0 ?

Answer 5

*did

Answer 6

sub in x=1? haha,

Answer 7

if u sub x = 1, u wud get 0/0 form , which is illegal.
so you cannot sub x =1 directly

Answer 8

heard of L'hospitals rule before ?

Answer 9

yeah thats the case, yup! were suppose to use that. so u mean just differentiate the whole expression? like,

\[g'(x)= \frac{ \frac{ 1 }{ x } }{ 1} \] Ah! now i get it, then how abt part b?

Answer 10

good :)

Answer 11

\(\large g'(1) = \lim \limits_{x \to 1} \frac{f(x) - f(1)}{x-1}\)

Answer 12

\(\large g'(1) = \lim \limits_{x \to 1} \frac{\frac{\ln x}{x-1} - 1}{x-1}\)

Answer 13

\(\large g'(1) = \lim \limits_{x \to 1} \frac{\ln x - x + 1}{(x-1)^2}\)

Answer 14

can u take it from here ?

Answer 15

yup! i think so, thanks! hm, i used this eqn\[\lim_{h \rightarrow 0} \frac{ g(1+h)-g(1) }{ h }\] so \[\lim_{h \rightarrow 0}\frac{ \frac{\ln(1+h) }{ (1+h)-1 }-1 }{h }\] i got h=-1/2

Answer 16

i think u used the other version of the definition

Answer 17

yes thats right !

Answer 18

yup both are one and same

Answer 19

okay thank u very much! :D you're great help indeed! (:

Answer 20

np :)