Help With Infinite Exponential Limit

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Homework Help Overview

The problem involves evaluating the limit of the expression \(\lim_{x \to - \infty} \frac{3^x-3^{-x}}{3^x+3^{-x}}\). The subject area is limits in calculus, particularly focusing on exponential functions as the variable approaches negative infinity.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss various algebraic manipulations to simplify the limit expression. There are questions about the steps taken in the original poster's solution, particularly regarding the transition between forms of the expression. Some participants suggest that understanding the behavior of exponential functions as \(x\) approaches negative infinity is crucial.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the limit and questioning the assumptions made in the calculations. Some guidance has been offered regarding the behavior of exponential terms, but no consensus has been reached on the correct evaluation of the limit.

Contextual Notes

There is a noted confusion regarding the limit's evaluation, with participants questioning the correctness of the original poster's conclusion that the limit equals -1, while they arrive at 1 instead. The discussion highlights the importance of understanding the limits of exponential functions as they approach negative infinity.

alba_ei
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Homework Statement


[tex] \lim_{x \to - \infty} \frac{3^x-3^{-x}}{3^x+3^{-x}}[/tex]

Homework Equations


[tex] \lim_{x \to - \infty} \frac{3^x-3^{-x}}{3^x+3^{-x}} = - 1[/tex]

The Attempt at a Solution


Im getting trouble when I try to evaluate this limit, altough the answer is -1 idont know how to get to it.

[tex] \lim_{x \to - \infty} \frac{3^x-3^{-x}}{3^x+3^{-x}}[/tex]

[tex] = \lim_{x \to - \infty} \frac{3^{-x}(3^{2x}-1)}{3^{-x}(3^{2x}+1)}[/tex]

[tex] = \lim_{x \to - \infty} \frac{3^{2x}-1}{3^{2x}+1}[/tex]

[tex] = \lim_{x \to - \infty} \frac{1-\frac{1}{3^{2x}}}{1+\frac{1}{3^{2x}}}[/tex]

[tex] = \lim_{x \to - \infty} \frac{1}{1} = 1[/tex]
I got 1, not -1
 
Last edited:
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How did you get +1?
 
you can do an algebraic trick to get that [itex]\frac{3^{x}-3^{-x}}{3^{x}+3^{-x}}=\frac{3^{2x}-1}{3^{2x}+1}[/itex] and then the limit is trivial.
 
I'm not sure how you got from the left hand side to the middle of your solution. Could you explain your steps further?
 
remember that when you take the limit of an exponential when the variable tends to [itex]-\infty[/itex] you get that this limit is 0. And remember that [itex]3^{2x}=e^{2x\ln{3}}[/itex]
 
Last edited:
Does 1/3^{2x} ---> 0 as x ---> -Infinity?
 
I think I messed up my reply by not quoting anyone, and now you've edited and put in more of your solution! Anyway...

alba_ei said:

Homework Statement


[tex] \lim_{x \to - \infty} \frac{3^x-3^{-x}}{3^x+3^{-x}}[/tex]

Homework Equations


[tex] \lim_{x \to - \infty} \frac{3^x-3^{-x}}{3^x+3^{-x}} = - 1[/tex]

The Attempt at a Solution


Im getting trouble when I try to evaluate this limit, altough the answer is -1 idont know how to get to it.

[tex] \lim_{x \to - \infty} \frac{3^x-3^{-x}}{3^x+3^{-x}}[/tex]

[tex] = \lim_{x \to - \infty} \frac{3^{-x}(3^{2x}-1)}{3^{-x}(3^{2x}+1)}[/tex]

[tex] = \lim_{x \to - \infty} \frac{3^{2x}-1}{3^{2x}+1}[/tex]

[tex] = \lim_{x \to - \infty} \frac{1-\frac{1}{3^{2x}}}{1+\frac{1}{3^{2x}}}[/tex]
There is no need to do this step. You should note that, since we are taking the limit x tends to negative infty, 3^2x tends to zero.
[tex] = \lim_{x \to - \infty} \frac{1}{1} = 1[/tex]
I got 1, not -1
 

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