Question

Find the limit without using L'Hopital's rule:
\[lim_{h\rightarrow 0} \frac{21.7^h-1}{h}\]

Answer 1

Took the ln of both sides. to separate, then took the limit, but I keep getting zero when I need 3.07731

Answer 2

@e.mccormick

Answer 3

if you could offer your valuable insight pretty please

Answer 4

OK, I can see where the 0 is coming from....
https://www.desmos.com/calculator/f95qx89kca

Answer 5

right, but i need 3.blahblah

Answer 6

Did you try separating the fraction into two fractions?

Answer 7

didn't help, I get something I can't apply a log to either

Answer 8

and evaluating the limit i get the first fraction minus infinity, which is not helpful

Answer 9

Yah, I was just noticing that... which is worse than getting 0!

Answer 10

Idk, where I went wrong lol indeed worse

Answer 11

i have this, \[ \lim_{h\rightarrow 0} ln(21.7^h-1)-ln(h)=ln(L)\]

Answer 12

after that take the lim, still get 0

Answer 13

you actually get 1 but

Answer 14

still wrong answer

Answer 15

you may want to mult top and bottom by \(27.1^h+1\)

Answer 16

ok let me give that a shot,

Answer 17

nah, not much help
I still can't get rid of the minus 1

Answer 18

what about the squeeze theorem?

Answer 19

nope, I don't know the answer from the get go

Answer 20

rewrite the limit.\[\lim_{h \rightarrow 0}\frac{ 27.1^h-1 }{ h }=\lim_{h \rightarrow 0}\frac{ e^{h\ln(27.1)}-1 }{ h }=\ln(27.1)\cdot \lim_{h \rightarrow 0}\frac{ e^{h\ln(27.1)}-1 }{ \ln(27.1)\cdot h }\]now let \(k=\ln(27.1)\cdot h\) then as \(h\rightarrow 0, k\rightarrow 0\) and you get
\[\ln(27.1)\cdot \lim_{h \rightarrow 0}\frac{ e^{h\ln(27.1)}-1 }{ \ln(27.1)\cdot h }=\ln(27.1)\cdot \lim_{k \rightarrow 0}\frac{ e^{k}-1 }{ k }=\ln(27.1)\]

Answer 21

hey @FibonacciChick666 , have a look

Answer 22

hmm last step, that's 0/0 l'hops gives you e^k-1=0 so stilll get 0 right?

Answer 23

i like the e^ln idea though

Answer 24

no, \[\lim_{h \rightarrow 0}\frac{ e^h-1 }{ h }=1\]by definition

Answer 25

really? I must be spacing sorry, how?

Answer 26

oh nvm i'm an idiot

Answer 27

you good?

Answer 28

yea ok, that works thank you

Answer 29

Idk why I couldn't think of anything

Answer 30

sorry, i forgot that anytime you have another base you always convert to base e (same goes for logs... always convert to ln).

Answer 31

yea, I got that, I just couldn't think of how to do this without l'hops

Answer 32

they haven't learned it yet

Answer 33

you can always check Paul's Online Math Notes. He's even got some nice "cheat sheets" there too!

Answer 34

yea, I know it. I'm just helping my bf, and for the life of me, couldn't figure this one out... calc was like 4 years ago O.O

Answer 35

have fun!

Answer 36

indeed. thanks again!