Question

Find the exact value of the following limit:
lim[x -> 0] (e*6x - 6x - 1)/(x^2)

Answer 1

L'Hopital's
twice

Answer 2

bdw, whats the numerator exactly?

Answer 3

\[\lim\limits_{x\to 0}\frac{e^{6x} - 6x-1}{x^2}\]

like this ?

Answer 4

But when i do the rule one i get -6/0 which doesn't meet the conditions

Answer 5

Yes, ganeshie8

Answer 6

hmm

(6e^(6x)-6) /2x

it is 0/0
how are u getting -6 ?

Answer 7

e^0=1

Answer 8

but when you take the derivative doesn't the 6x turn into 6

Answer 9

wait never mind i see where i made my mistake

Answer 10

sure

\(\Large (e^{6x})' = 6e^{6x} \\ \large (-6x)' =-6\)

Answer 11

any more doubts? :)

Answer 12

No more doubts i know how to do it now

Answer 13

^_^

Answer 14

Why don't you guys try it out the hard way. It's actually better learning

Answer 15

whats the hard way ?

Answer 16

the definition is bretty much easy :P

Answer 17

Without Hopital's rule. Analytical solving..

Answer 18

epsilon delta ? for that you need to know the limit upfront right ?

Answer 19

ohk..

Answer 20

the numerator ain't factorable

Answer 21

Thank you guys so much for your help. XD

Answer 22

taylor series is also a kindof lhopital, so you dont want to use it either ?

Answer 23

this may be useful\[\large \lim\limits_{x\to 0}\dfrac{e^x-1}{x} = 1\]

Answer 24

exactly

Answer 25

the denominator is x^2 however, so im not so sure

Answer 26

This is a really odd way... but \[u=x^2 \text{ as } x->0, u->0 \\ \pm \sqrt{u}=x \text{ so we have two cases to work } \\ \text{ using } x=\sqrt{u} \text{ we have } \\ f'(x^+)=f'(\sqrt{u})=\lim_{u \rightarrow 0} \frac{f(u)-f(0)}{u-0} \\ f(u)=e^{6 \sqrt{u}}-6 \sqrt{u} \\ f'(u)=\frac{6}{2 \sqrt{u}}e^{6 \sqrt{u}}-\frac{6}{2 \sqrt{u}} =\frac{3}{\sqrt{u}}e^{6 \sqrt{u}}-\frac{3}{\sqrt{u}}=\frac{3e^{6 \sqrt{u}}-3}{\sqrt{u}}=3 \frac{e^{6 x}-1}{x} \\ \text{ let } v=6x \\ \lim_{x \rightarrow 0^+} 3 \frac{e^{6x}-1}{x} = \lim_{v \rightarrow 0^+} 3 \frac{e^{v}-1}{\frac{v}{6}}=18(1)=18\]
\[f'(x^-)=f'(- \sqrt{u}) \\ f(u)=e^{-6 \sqrt{u}}+6 \sqrt{u} \\ f'(u)=\frac{-6}{2 \sqrt{u}}e^{-6 \sqrt{u}} +\frac{6 }{2 \sqrt{u}} =-3\frac{ e^{ -6\sqrt{u}}-1}{\sqrt{u}}=3 \frac{e^{6x}-1}{x}\]
so we still get the same limit 18
This is a little odd but it seems to work

Answer 27

thank goodness there is L'Hopital :O

Answer 28

I don't know if that is the way the others were talking of or not

Answer 29

nice :)