Question

Solve limit using Hospital rule.

lim x-> -infinity ( (2^x-2^(-x)) / 2^x+2^(-x) )

The answer from WolphramAlpha is -1, but how?

Answer 1

I got this so far:

\[\frac{ 2^{x}\ln(2)+2^{-x}\ln(2) }{ 2^{x}\ln(2)-2^{-x}\ln(2) } \]

How do I proceed?

Answer 2

I'm wondering why we need l'Hopital.
Perhaps pondering the limit as \(x \rightarrow \infty\) might contribute to understanding.

Answer 3

I think it's because it's currently in the form of infinity/infinity, which is an indeterminate form.

Answer 4

Can we write it like this?
\[\LARGE \frac{4^x-1}{4^x+1}\]

Answer 5

x->infinity

Answer 6

@continuume

Answer 7

How did you get that?

Answer 8

\[\LARGE \lim_{x \rightarrow \infty } \frac{2^x-2^{-x}}{2^{-x}+2^x}\]
right?

Answer 9

Yes

Answer 10

Wait, to negative infinity

Answer 11

yes,typo^^

Answer 12

Try to simplify this by yourself once,just simplify

Answer 13

I did, which is what I posted earlier up. But then I don't know how to proceed with that..

Answer 14

Okay

Answer 15

\[\LARGE \lim_{x \rightarrow \infty} \frac{2^{x}-\frac{1}{2^x}}{\frac{1}{2^{x}}+2^{x}}\]

Answer 16

can u do it now?

Answer 17

\[\LARGE 2^{-x}=\frac{1}{2^{x}}\]
u know this right?

Answer 18

Yes, I can now see how your answer came about.

Answer 19

Glad,lets proceed.

Answer 20

\[\LARGE \lim_{x \rightarrow \infty} \frac{4^{x}-1}{4^{x}+1}\]

Answer 21

thats -infty sorrry dam!

Answer 22

Lol

Answer 23

I see the answer now!! :)

Answer 24

\[\LARGE \lim_{x \rightarrow -\infty} \frac{4^{x}}{1+4^{x}} -\lim_{x \rightarrow -\infty} \frac{1}{1+4^{x}}\]
can i write like this?

Answer 25

Yes

Answer 26

u got the answer?

Answer 27

I don't know if I did it write, but I did 4 to the power of infinity, which yields 0. I don't know if that's right.

Answer 28

Simplifying, I get -1.

Answer 29

okay it must be correct then :P

Answer 30

\[\frac{ 4^{\infty}-1 }{ 4^{\infty}+1 } = -1\]

Not sure if correct ... :p

Answer 31

http://www.wolframalpha.com/input/?i=%284%5Einfinity-1%29%2F%284%5Einfinity+%2B1%29
Indeterminate,sir.

Answer 32

I get -1 from wolphramAlpha, hmm...

Answer 33

:S

Answer 34

lets move ahead?

Answer 35

Actually you're right,....it's indeterminate :/

Answer 36

:)

Answer 37

You should know,limit of a constant is constant..
and limit of a quotient is the quotient too!

Answer 38

\[\LARGE \frac{1}{\frac{1}{{\lim_{x \rightarrow -\infty}4^{x}}} +1} - \frac{1}{{\lim_{x \rightarrow -\infty}1+4^{x}}} \]

Answer 39

now use continuity

Answer 40

why dont u apply L hospital after the 2nd step :P

Answer 41

o.o sensei's here!! :D hey @zepdrix

Answer 42

he's gone :(

Answer 43

nvm

Answer 44

sensei hai, imasu o.o

Answer 45

konichiwa. I'm still a bit lost..

Answer 46

\[\huge \text{:O}\]

Answer 47

WHY U NO FOLLOW WHAT I SAY

Answer 48

(╯°□°）╯︵ ┻━┻)

Answer 49

Woah woah woah woah! Simmer down now!
┬─┬ ノ( ゜-゜ノ)