Question

find
lim x->0 ln(x+1)/x

using the derivative

Answer 1

we have
\[\lim_{x\to 0}\frac{\ln (x+1)}{x}\]
if we put x=0 then we get
\[\frac 0 0\] so we can apply L'Hopitol here. Take derivative of both numerator and denominator, we get
\[\huge \lim_{x\to 0}\frac{\frac 1{(x+1)}}{1}\]
can you solve now?

Answer 2

wait a minute i think i found the solution for this, recalling some stuff from pre-calculus

Answer 3

*L'Hopital 's

Answer 4

cant't use L'Hopsital rule

Answer 5

i know i s 1

Answer 6

you mentioned that it has to be done using the derivative

Answer 7

correct

Answer 8

I did the same thing

Answer 9

but forget to mention that u can't use L'Hosp for this proble

Answer 10

i recall that ln (m/n ) u can change to ln m- ln n

Answer 11

i guess this problem is involve with chain rule

Answer 12

I got something

Answer 13

We know
\[f'(x)=\lim_{h\to 0} \frac{f(x+h)-f(x)}{x+h-x}\]

Answer 14

k that may work

Answer 15

let
\[f(x) =\ln (x+1)\]
so
\[f'(x)=\lim_{h\to 0} \frac{\ln (x+h+1)-\ln(x+1)}{x+h-x}\]
we know that f'(x)= 1/(x+1 )so
\[\frac{1}{x+1}=\lim_{h\to 0} \frac{\ln (x+h+1)-\ln(x+1)}{x+h-x}\]
\[\large \frac{1}{x+1}=\lim_{h\to 0} \frac{\ln (\frac{x+h+1}{x+1})}{h}\]
we get
\[\large \frac{1}{x+1}=\lim_{h\to 0} \frac{\ln (\frac{h}{x+1}+1)}{h}\]
put x=0
\[\large \frac{1}{1}=\underline{\lim_{h\to 0} \frac{\ln (\frac{h}{1}+1)}{h}}\]
this is our limit so, we have proved that it's equal to 1
just replace h by x, cause it doesn't depend on variable

Answer 16

@Andresfon12 do you understand this?

Answer 17

yes @ash2326

Answer 18

good :)

Answer 19

very clever

Answer 20

nice explanation @ash2326 ..