Is there a condition for applying standard limits?

  • #1
Mcp
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Main Question or Discussion Point

I came across this basic limits question

Ltx->0[(ln(1+X)-sin(X)+X2/2]/[Xtan(X)Sin(X)]

The part before '/'(the one separated by ][ is numerator and the one after that is denominator

The problem is if I substitute standard limits :
(Ltx->0tan(X)/x=1
Ltx->0sin(X)/X=1
Ltx->0ln(1+X)/X=1)
The expression will simplify to
Ltx->0[X2/2]/[X]3
=Ltx->01/2X
Which is not defined

But if I had only put Ltx->0sin(X)=X and Ltx->0tan(X)=X in the denominator, the denominator would become [X]3 and used polynomial expansion of ln(1+X) and sin(X) in numerator, the coefficient of [X]3 in the numerator would have been the answer (as smaller power terms cancel out and larger ones tend to 0) which here is 1/2, the correct answer.

So why didn't I get the right answer using standard limits.

Regards
 
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Answers and Replies

  • #2
Math_QED
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I have a very hard time reading your post. Is it possible to format it in Latex?
 
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  • #4
RPinPA
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I'm going to quote your entire post as an opportunity to fix the formatting. Meanwhile, I recommend you follow people's suggestions and experiment with the LaTeX formatting to get a better chance of answers in future.

Also, please don't use both ##X## and ##x## interchangeably. They are two different symbols, so your expressions are much more readable if you just stick to one.

I came across this basic limits question

$$\lim_{x \rightarrow 0} \frac {\ln(1+x)−\sin(x)+\frac{x^2}{2}} {xtan(x)sin(x)}$$​

The problem is if I substitute standard limits :
$$\lim_{x \rightarrow 0} \frac {\tan(x)} {x} = 1 \\
\lim_{x \rightarrow 0} \frac {\sin(x)} {x} = 1 \\
\lim_{x \rightarrow 0} \frac {\ln(1+x)} {x} = 1$$

The expression will simplify to
$$\lim_{x \rightarrow 0} \frac {\left ( \frac {x^2}{2} \right )} {x^3} = \lim_{x \rightarrow 0} \frac {1}{2x}$$

Which is not defined
I'm having a little trouble following that calculation. Can you break it down in smaller steps?

But if I had only put ##\lim_{x→0} _sin(x)=x and ##\lim_{x→0} tan(x)=x in the denominator,
Technically those aren't the limits, but approximations for small xx. Because the error in the approximations goes to 0 as ##x \rightarrow 0##, they could be used as the basis of a limit argument.

the denominator would become ##x^3## and used polynomial expansion of ##\ln(1+x)## and ##\sin(x)## in numerator, the coefficient of ##x^3## in the numerator would have been the answer (as smaller power terms cancel out and larger ones tend to 0) which here is 1/2, the correct answer.

So why didn't I get the right answer using standard limits.
I have a feeling there's an error, possibly subtle, in how you used the "standard limits". Could you expand on that part of the calculation please?
 
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  • #5
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I have a very hard time reading your post.
So do I. Besides that difficulty, I'm not sure what you mean by this part, the numerator in your limit expression, and particularly what gets divided by 2.
[(ln(1+X)-sin(X)+X2/2]
Is this what you intend?
##[\ln(1 + x)\sin(x) + \frac{x^2}2]##
Or is it this?
##[\frac{\ln(1 + x)\sin(x) + x^2}2]##
 
  • #6
pasmith
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I came across this basic limits question

Ltx->0[(ln(1+X)-sin(X)+X2/2]/[Xtan(X)Sin(X)]

The part before '/'(the one separated by ][ is numerator and the one after that is denominator

The problem is if I substitute standard limits :
(Ltx->0tan(X)/x=1
Ltx->0sin(X)/X=1
Ltx->0ln(1+X)/X=1)
The expression will simplify to
Ltx->0[X2/2]/[X]3
=Ltx->01/2X
Which is not defined

But if I had only put Ltx->0sin(X)=X and Ltx->0tan(X)=X in the denominator, the denominator would become [X]3 and used polynomial expansion of ln(1+X) and sin(X) in numerator, the coefficient of [X]3 in the numerator would have been the answer (as smaller power terms cancel out and larger ones tend to 0) which here is 1/2, the correct answer.

So why didn't I get the right answer using standard limits.

Regards
To leading order, the numerator is [tex]\ln(1 + x) - \sin x + \tfrac12 x^2 = \left(x - \tfrac12 x^2 + \tfrac13 x^3 + O(x^4) \right) - \left(x - \tfrac{x^3}{6} + O(x^5)\right) + \tfrac12 x^2 = \tfrac12 x^3 + O(x^4).[/tex] You have somehow concluded that it was [itex]\frac12x^2 + O(x^3)[/itex], but as you haven't shown any working we can't tell you where you went wrong.

Bear in mind that as this is an indeterminate form of type 0/0 it is not legitimate to split the expression into three fractions, take the limits separately, and then add them together; you must keep it as a single fraction.
 
  • #7
RPinPA
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Aargh! Every time I edit my own post, it screws up the formatting and deletes a bunch of the LaTeX!

Managed to fix most of it, but I'm not touching that post again as I'll lose all the formatting again if I try.
 
  • #8
Mcp
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To leading order, the numerator is [tex]\ln(1 + x) - \sin x + \tfrac12 x^2 = \left(x - \tfrac12 x^2 + \tfrac13 x^3 + O(x^4) \right) - \left(x - \tfrac{x^3}{6} + O(x^5)\right) + \tfrac12 x^2 = \tfrac12 x^3 + O(x^4).[/tex] You have somehow concluded that it was [itex]\frac12x^2 + O(x^3)[/itex], but as you haven't shown any working we can't tell you where you went wrong.

Bear in mind that as this is an indeterminate form of type 0/0 it is not legitimate to split the expression into three fractions, take the limits separately, and then add them together; you must keep it as a single fraction.
The answer using polynomial expansion of ln(1+X) and sin(X) in numerator is 1/2 as numerator is X3/2 + higher powers of X, which(those higher powers) tend to 0 and by putting tan(X)=X and sin(X)=X in the denominator, the denominator becomes X3 which gives the right answer.

But I am asking why cant I substitute ln(1+X) and -sin(X) in the numerator by X and -X respectively as X is very small. If I were to do so then only X2/2 would be left in the numerator while X3 in the denominator which of course will not yield 1/2.

Sorry about the LaTeX thing I don't know much about it.
 
  • #9
pasmith
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But I am asking why cant I substitute ln(1+X) and -sin(X) in the numerator by X and -X respectively as X is very small. If I were to do so then only X2/2 would be left in the numerator while X3 in the denominator which of course will not yield 1/2.
The [itex]O(x)[/itex] terms from the log and sine expansions cancel. You can't conclude from that that the rest of the terms in the expansions will cancel, which is effectively what you do by ignoring them because they are "small". They may be small, but they are the largest terms which don't vanish for [itex]x \neq 0[/itex].

So you have to look next at the [itex]O(x^2)[/itex] terms. The sine expansion doesn't have one, but the log expansion does and it is cancelled by the [itex]\frac12 x^2[/itex] term. So you have then to look at the [itex]O(x^3)[/itex] terms, and those don't cancel but sum to [itex]\frac12 x^3.[/itex]
 
  • #10
RPinPA
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For one thing, because the linear terms cancel for all x in the radius of convergence, and therefore for small ##x## the value of ##\ln(1 + x) - \sin(x)## is given by the higher order terms. I assumed you were considering the higher order terms when you talked about the "polynomial expansion". You need to, to understand the behavior of ##\ln(1 + x) - \sin(x) + (x^2/2)## at small x.
 
  • #11
Math_QED
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The answer using polynomial expansion of ln(1+X) and sin(X) in numerator is 1/2 as numerator is X3/2 + higher powers of X, which(those higher powers) tend to 0 and by putting tan(X)=X and sin(X)=X in the denominator, the denominator becomes X3 which gives the right answer.

But I am asking why cant I substitute ln(1+X) and -sin(X) in the numerator by X and -X respectively as X is very small. If I were to do so then only X2/2 would be left in the numerator while X3 in the denominator which of course will not yield 1/2.

Sorry about the LaTeX thing I don't know much about it.
There isn't much to know. It takes 5 minuts to learn it. Just putting your equations between dollar signs or # signs will do wonders:

Example:

Instead of 1/(x^2+1), type 1/(x^2+1) with two dollar signs or two hastags left and right from it. It will look like this: ##1/(x^2+1)##. I invite you to experiment with this. You might need it anyway sooner or later.
 
  • #12
Stephen Tashi
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So why didn't I get the right answer using standard limits.
Apparently, what you mean by "using standard limits" is assuming things like:

If ##lim_{x \rightarrow a} g(x) = L## then ##lim_{x \rightarrow a} f(g(x)) = f(L)##

and

If ##lim_{x \rightarrow a} g(x) = L_1## and ##lim_{x \rightarrow a} h(x) = L_2## then ##\lim_{x \rightarrow a} f(g(x), h(x)) = f(L_1, L_2)##


Those statements are not, in general, true. Of course, in many practical problems, we encounter functions where those ideas do work. So if you don't study the theory behind limits and only go by examples, you can get the impression that such ideas are 100% reliable.

To get the theory straight, you should read carefully the theorems that add more conditions to the "if" part of those ideas and are 100% reliable.
 

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