Does anyone know the best way to show this limit does not exist?

  • Thread starter AxiomOfChoice
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In summary: So we can indeed say that \lim_{x\to a} \Big( f(x) + g(x) \Big) = \infty. In summary, the most efficient way to show that the limit in question does not exist is by using Taylor expansions and the fact that the left and right-sided limits of the derivatives are infinite. L'Hopital's rule can also be used, but only if it can be shown that the left and right limits of the derivatives are infinite. However, it may not be applicable if the limits of the individual functions are infinite and have different signs.
  • #1
AxiomOfChoice
533
1
What's the most efficient way (other than just plotting it and eyeballing it) to show that

[tex]
\lim_{x\to 0} \frac{\ln (1-x) - \sin x}{1 - \cos^2 x}
[/tex]

does not exist?
 
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  • #2
l'Hôpital?
 
  • #3
tiny-tim said:
l'Hôpital?



Can't be. L'Hospital gives a sufficient condition for a limit to exist, not necessary, meaning: if the limit *after* doing L'H doesn't exist we cannot deduce the original limit doesn't exist either.

DonAntonio
 
  • #4
DonAntonio said:
Can't be. L'Hospital gives a sufficient condition for a limit to exist, not necessary, meaning: if the limit *after* doing L'H doesn't exist we cannot deduce the original limit doesn't exist either.

DonAntonio

Still, we can use L'Hopitals rule by noting it holds also for one-sided limits and for extended limits. So if we can show that the left-sides and right-sides limits of the derivatives are infinite, then so are the left-sided and rigt-sided limits of the original function.
 
  • #5
AxiomOfChoice said:
What's the most efficient way (other than just plotting it and eyeballing it) to show that

[tex]
\lim_{x\to 0} \frac{\ln (1-x) - \sin x}{1 - \cos^2 x}
[/tex]

does not exist?



One way is using Taylor expansions: [tex]\log(1-x)=-(x+\frac{x^2}{2}+\frac{x^3}{3}+...)\,\,,\,\,\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-...\Longrightarrow \sin^2x=x^2-\frac{x^4}{3}+...[/tex] so[tex]\frac{\log(1-x)\sin x}{\sin^2x}=\frac{-2x-\frac{x^2}{2}-\frac{x^3}{6}-...}{x^2-\frac{x^4}{3}+...}=-\frac{1}{x}\frac{2+\frac{x}{2}+\frac{x^2}{3}+...}{1-\frac{x^2}{3}+...}[/tex]

The second factor in the last expression above has limit [itex]\,2\,\text{when}\,x\to 0\,[/itex] , whereas the first factor has no limit, as it approaches [itex]\,-\infty\,\,\,or\,\,\,+\infty\,[/itex] according as whether [itex]\,x\to 0^+\,\,\,or\,\,\,x\to 0^-\,[/itex] , resp.

DonAntonio
 
  • #6
[tex]\frac{\ln(1-x)-\sin x}{1-\cos^2x} = \frac{\ln(1-x)-\sin x}{\sin^2x} = \frac{\ln(1-x)}{\sin^2x} - \csc x[/tex]
[tex]\lim_{x\to 0^-} \frac{\ln(1-x)}{\sin^2x} = +\infty[/tex] [tex]\lim_{x\to 0^+}\frac{\ln(1-x)}{\sin^2x} = -\infty[/tex][tex]\lim_{x\to 0^-}-\csc x = +\infty[/tex][tex]\lim_{x\to 0^+}-\csc x = -\infty[/tex]

The left and right-sided limits aren't the same, so the limit doesn't exist.
 
  • #7
micromass said:
Still, we can use L'Hopitals rule by noting it holds also for one-sided limits and for extended limits. So if we can show that the left-sides and right-sides limits of the derivatives are infinite, then so are the left-sided and rigt-sided limits of the original function.

Does this mean that L'Hopital's rule is still good even if we rewrite the statement of the theorem in terms of left-hand (or right-hand) limits? (We obviously then just obtain a result about the left-hand or right-hand limit of the function.)
 
  • #8
Bohrok said:
[tex]\frac{\ln(1-x)-\sin x}{1-\cos^2x} = \frac{\ln(1-x)-\sin x}{\sin^2x} = \frac{\ln(1-x)}{\sin^2x} - \csc x[/tex]
[tex]\lim_{x\to 0^-} \frac{\ln(1-x)}{\sin^2x} = +\infty[/tex] [tex]\lim_{x\to 0^+}\frac{\ln(1-x)}{\sin^2x} = -\infty[/tex][tex]\lim_{x\to 0^-}-\csc x = +\infty[/tex][tex]\lim_{x\to 0^+}-\csc x = -\infty[/tex]

The left and right-sided limits aren't the same, so the limit doesn't exist.
You seem to be assuming that we can determine

[tex]
\lim_{x\to a} \Big( f(x) + g(x) \Big)
[/tex]

by considering

[tex]
\lim_{x\to a} f(x) + \lim_{x\to a} g(x),
[/tex]

even if the limits of [itex]f[/itex] and [itex]g[/itex] are [itex]\pm \infty[/itex]. Is this really the case?
 
  • #9
AxiomOfChoice said:
Does this mean that L'Hopital's rule is still good even if we rewrite the statement of the theorem in terms of left-hand (or right-hand) limits?

yes … if we can use l'hôpital to prove that the left and right limits are ∞ and -∞, then we have proved that "the" limit does not exist :wink:
 
  • #10
AxiomOfChoice said:
Does this mean that L'Hopital's rule is still good even if we rewrite the statement of the theorem in terms of left-hand (or right-hand) limits? (We obviously then just obtain a result about the left-hand or right-hand limit of the function.)

A function has a limit at x0 iff its right and left limits at x0 exist and are equal.
 
  • #11
[tex]
\lim_{x\to 0} x \frac{\ln (1-x) - \sin x}{1 - \cos^2 x}=-2
[/tex]
 
  • #12
lurflurf said:
[tex]
\lim_{x\to 0} x \frac{\ln (1-x) - \sin x}{1 - \cos^2 x}=-2
[/tex]
Nope. That's [itex]\lim_{x\to 0} x \frac{\ln (1-x) - \sin x}{\sin x}[/itex]. It looks like you lost a power of 2 in the denominator.
 
  • #13
AxiomOfChoice said:
You seem to be assuming that we can determine

[tex]
\lim_{x\to a} \Big( f(x) + g(x) \Big)
[/tex]

by considering

[tex]
\lim_{x\to a} f(x) + \lim_{x\to a} g(x),
[/tex]

even if the limits of [itex]f[/itex] and [itex]g[/itex] are [itex]\pm \infty[/itex]. Is this really the case?

It depends on whether the infinite limits have the same sign. If, say, [itex]\lim_{x\to a} f(x) = \infty[/itex] while [itex]\lim_{x\to a} g(x) = -\infty[/itex], then we have to use other methods to investigate the limit. But, informally speaking, if [itex]\lim_{x\to a} f(x) = \lim_{x\to a} g(x) = \infty[/itex] (note: same sign), then close to a, f(x) is very large and g(x) is nonnegative, which means that f(x)+g(x)≥f(x)+0, which is still very large.
 

1. What is a limit in mathematics?

A limit in mathematics refers to the value that a function approaches as the input approaches a certain value. It is used to describe the behavior of a function near a specific point and can help determine if a function is continuous or discontinuous.

2. How do you show that a limit does not exist?

To show that a limit does not exist, you can use different methods such as the ε-δ definition, the squeeze theorem, or by finding different left and right limits. Ultimately, you need to find a contradiction or inconsistency in the limit definition to prove that it does not exist.

3. What are some common reasons for a limit to not exist?

A limit may not exist if the function has a vertical asymptote, a jump discontinuity, or an infinite oscillation near the point of interest. It can also not exist if there are different left and right limits at the point, or if the function is not defined at that point.

4. Can you give an example of a limit that does not exist?

One example of a limit that does not exist is the function f(x) = 1/x as x approaches 0. As x gets closer to 0 from the left, the function approaches negative infinity, while as x gets closer to 0 from the right, the function approaches positive infinity. Therefore, the limit does not exist at x = 0.

5. How can knowing that a limit does not exist be useful in mathematical applications?

Knowing that a limit does not exist can help identify points of discontinuity in a function, which can be important in real-life applications such as engineering or physics. It can also help determine the behavior of a function near a specific point, which can be useful in optimization and curve fitting problems.

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