Limits to Infinity: Exponential Function

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    Infinity Limits
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Discussion Overview

The discussion revolves around evaluating limits involving exponential functions and rational expressions as the variable approaches infinity or negative infinity. Participants explore various limit problems, including the application of L'Hôpital's Rule, the behavior of functions at infinity, and the use of algebraic manipulation to determine limits.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested
  • Homework-related

Main Points Raised

  • One participant proposes a method for evaluating limits of exponential functions by first determining the limit of the argument of the exponential function.
  • Another participant confirms that the limit of the argument can be substituted into the exponential function to find the limit.
  • Discussion includes evaluating limits as x approaches negative infinity, with one participant suggesting that e raised to negative infinity approaches 0.
  • Participants discuss the limit of a rational expression as x approaches negative infinity, with differing views on how to handle the expression -∞/∞.
  • Some participants suggest that when the highest powers in the numerator and denominator are the same, the limit can be found by dividing the coefficients of those powers.
  • One participant mentions using L'Hôpital's Rule for certain limits, while others express a preference for alternative methods.
  • Challenges are introduced, with participants sharing specific limit problems that require more thought, including the use of the squeeze theorem and known limits.
  • There is a mention of the terms "abscissa" and "ordinate" in relation to ordered pairs, indicating a side discussion on terminology.

Areas of Agreement / Disagreement

Participants generally agree on some methods for evaluating limits, but there are multiple competing views on the best approaches, particularly regarding the handling of indeterminate forms and the use of L'Hôpital's Rule. The discussion remains unresolved on certain limit evaluations and methods.

Contextual Notes

Some participants express uncertainty about the application of certain limit techniques and the appropriateness of specific algebraic manipulations. There are also references to the limitations of memorization in calculus versus understanding underlying concepts.

Who May Find This Useful

Students and enthusiasts of calculus, particularly those interested in limit evaluation techniques and the behavior of functions at infinity.

  • #31
Four is killing me...

Confunction identities? <- nvm don't think that's it
 
Last edited:
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  • #32
multiply numerator and denominator by cos(x)+1.
 
  • #33
Ohhhhhhh, I tried that, but never thought about plugging in 0...fml

sin(x)/(cos(x)+1) ->

0/2=0

....wow (again)
 
  • #34
solution to 2 without using hospital or Taylor:
limx→inf (1+1/x)^x = e
<=> limx→inf xln(1+1/x) =1
<=> limx→0 ln(1+x)/x = 1
<=> limx→0 x - ln(1+x) = 0
<=> limx→0 (ex-1)/x = e^0 =1
 
  • #35
NeroKid said:
solution to 2 without using hospital or Taylor:
limx→inf (1+1/x)^x = e
<=> limx→inf xln(1+1/x) =1
<=> limx→0 ln(1+x)/x = 1
<=> limx→0 x - ln(1+x) = 0
<=> limx→0 (ex-1)/x = e^0 =1

I'm not quite following you. I know that (1+1/n)n = e as n->infinity
 
  • #36
u take the ln at the both side of limx --> inf (1+1/x)^x = e
and from there u switch to x' which equal to 1/x
 
  • #37
How did you get from here:
<=> limx→inf xln(1+1/x) =1
to here?:
<=> limx→0 ln(1+x)/x = 1
 
  • #38
limx→inf xln(1+1/x) =1
<=> lim(1/x)→0 xln(1+1/x)=1
call x' = 1/x
limx'→0 ln(1+x')/x' =1
 
  • #39
NeroKid said:
limx→inf xln(1+1/x) =1
<=> lim(1/x)→0 xln(1+1/x)=1
call x' = 1/x
limx'→0 ln(1+x')/x' =1

Ok, I see.
Thanks
 
  • #40
Lastly,

How did you get these two steps?

<=> limx→0 ln(1+x')/x' = 1
<=> limx→0 x - ln(1+x) = 0
 
Last edited:
  • #41
because limx→0 ln(1+x)/x = 1 and limx→0 ln(1+x) = 0 and limx→0 x =0
so limx→0 ln(1+x) = limx→0 x
 
  • #42
Got it. Thanks
 

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