Question

Find the limit. Use L'Hopital's rule if it applies.
lim [1-cosh (8x)]/x^2 =???
x → 0

Answer 1

any help would be appreciated!! thank you!!!

Answer 2

so does l'hosptal apply do we have 0/0?

Answer 3

by the way \[\cosh (8x)=\frac{e^{8x}+e^{-8x}}{2}\]
if that helps you decide if l'hosptial applies

Answer 4

so its 2/2?

Answer 5

which is 1?

Answer 6

and we would then have (1-1)/0^2
so we do have 0/0 :)

Answer 7

and that means we can apply l'hospital's rule

Answer 8

yes that's what i have too in one of my steps, but the limit i got once having applied the rule was 0 but my hw said to was wrong ;/

Answer 9

it*

Answer 10

do you prefer to leave cosh(8x) or can we write it in terms of those exponential functions
your call
i prefer the exponential function way because I'm actually less familiar with all the properties derived from the hyperbolic functions

Answer 11

yeah the exponential is fine :)

Answer 12

\[\lim_{x \rightarrow 0}\frac{1-\frac{e^{8x}+e^{-8x}}{2}}{x^2}=\lim_{x \rightarrow 0}\frac{2-(e^{8x}+e^{-8x})}{2x^2} =\lim_{x \rightarrow 0} \frac{2-e^{8x}-e^{-8x}}{2x^2}\]

Answer 13

ok so this is the function we have

Answer 14

the derivative of the bottom is pretty easy...

Answer 15

so let's just jump straight to the top

Answer 16

sounds great

Answer 17

what is the derivative of the top ?

Answer 18

so its -8e^8x+8e^8x?

Answer 19

for the top

Answer 20

\[\lim_{x \rightarrow 0}\frac{0-8e^{8x}+8e^{-8x}}{4x}\]
well that one exponent show still be negative
but other than that great :)

Answer 21

oops yeah i always lose my signs ;/ lol. so now we just plug in the limit for each and see what we get?

Answer 22

yep and to our surprise we have ?

Answer 23

drum rolls

Answer 24

the bottom limit would be 0 so wouldn't that automatically make it 0?

Answer 25

well both top and bottom equals 0

Answer 26

0/0
so we have to do l'hospital again

Answer 27

but don't worry this is the last time

Answer 28

lol okay lets see

Answer 29

I know you can do the bottom

Answer 30

try differentiating that top again

Answer 31

well the new top we have

Answer 32

okay hang on

Answer 33

k hanging

Answer 34

-64e^-8x(e^(16x)+1 :S lol

Answer 35

ok ummm...
well hmmm...
let me help you with that

Answer 36

\[(-8e^{8x}+8e^{-8x})' \\ -8(8x)'e^{8x}+8(-8x)'e^{-8x} =?\]

Answer 37

-64xe^(8x)-64xe^(-8x)

Answer 38

tons better

Answer 39

\[\lim_{x \rightarrow 0}\frac{-64e^{8x}-64e^{-8x}}{4}\]

Answer 40

lol so now i just take the limit as x-->0?

Answer 41

see how we want use l'hosptial again because the bottom is 4

Answer 42

right

Answer 43

this is the final step

Answer 44

just plug in 0

Answer 45

ohhhokay that makes tons of sense

Answer 46

won't* (not want)

Answer 47

alrighty! thank you soooo much!!! you were a great help!!1

Answer 48

thanks

Answer 49

and np