Why Does Splitting Limits Cause Problems with 0/0 Forms?

In summary: The Limit Laws allow for factoring out of limits when both factors converge. When sin/sin=1, this is the case because the limits are getting closer and closer to each other. However, when sin/sin is undefined, as in the case with the limit 0/0, the limit laws can't help us because the limits are not converging.
  • #1
opus
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Homework Statement


Evaluate: $$\lim_{θ \rightarrow 0} {\frac{1-cos θ}{sin θ}}$$

Homework Equations

The Attempt at a Solution



By using trigonometric identities, I get to:

$$\lim_{θ \rightarrow 0} {\frac{sin θ}{sin θ}}⋅\lim_{θ \rightarrow 0} {\frac{sin θ}{1+cos θ}}$$

By using the Limit Laws, I can evaluate the limit of the numerator and denominator of each factor individually, then multiply these limits together to get the limit of the given function:

$$\lim_{θ \rightarrow 0} {\frac{sin θ}{sin θ}} = \frac{0}{0}$$
$$\lim_{θ \rightarrow 0} {\frac{sin θ}{1+cos θ}}=\frac{0}{2}=0$$

Multiplying the two limits together, ##\frac{0}{0}⋅\frac{0}{2}=0##

Here is my confusion:

For ##\lim_{θ \rightarrow 0} {\frac{sin θ}{sin θ}}##, I am looking at this two ways. One is that I can use Limit Laws to evaluate the numerator and denominator separately, and their quotient is the limit of the whole thing. This gives ##\frac{0}{0}## which is undefined so that makes no sense.
But if I look at it as ##\frac{sin θ}{sinθ}=1##, this leaves me with 1⋅0=0 for the solution to the whole problem which does make sense.
So what's the deal with the Limit Law problem? Why do I get an undefined limit using them, whereas if I just take sin/sin=1, I get a defined answer?
 
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  • #2
As you have been told in other threads, you cannot go around just factoring things out of limits. you can only do that when both factors converge nicely. In particular you should never ever ever factor something out of a limit if you are not sure it converges. Obviously 0/0 does not make sense at all. That is why you need the limit in the first place.

Also, what was wrong with just cancelling the sines? The argument is going to zero, it is not zero.
 
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  • #3
Orodruin said:
As you have been told in other threads, you cannot go around just factoring things out of limits. you can only do that when both factors converge nicely.

Isn't the purpose of the Limit Laws to be able to take bigger functions or expressions, and break them down into more manageable things i.e factoring things out? And I'm not sure what you mean by "converges". I haven't heard that term yet. Does it mean that the limit is the same as the x value gets approached from both sides?

Orodruin said:
Also, what was wrong with just cancelling the sines?
Nothing is wrong with it (I wouldn't think). But because I want to understand the concepts, I have to ask questions such as this.

Orodruin said:
The argument is going to zero, it is not zero.
That is very helpful and I suppose I didn't make that distinction. So then let me ask this- for the second part, which is ##\frac{0}{2}##, this isn't saying that the function is equal to zero, but that it's approaching it. Now how does this fit into the picture of the Limit Laws? If say limit ##L=1## and limit ##M=\frac{0}{2}##, the limit laws say that I can multiply these two limits together to get the limit as a whole. But since these are not necessarily values that the function is equal to at that point x, how can we multiply them together if they're "not equal to zero" but going to zero instead? In other words, we aren't multiplying any number by any number, but a series of numbers to a series of numbers.
 
  • #4
opus said:
the limit laws say that I can multiply these two limits together to get the limit as a whole.

Both 1 and 0/2 are well defined, so it makes sense to multiply them. But 0/0 is not well defined, it is undefined. Depending on the original function, the actual behavior as x approaches zero could be very different than what you were expecting. So trying to multiply 0/0 with something is problematic since 0/0 isn't defined in the first place. You can't get a single value as an answer (in other words ##\frac{0}{0}*\frac{0}{2}≠0##, it's an invalid operation).
 
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  • #5
Drakkith said:
Both 1 and 0/2 are well defined, so it makes sense to multiply them. But 0/0 is not well defined, it is undefined. Depending on the original function, the actual behavior as x approaches zero could be very different than what you were expecting. So trying to multiply 0/0 with something is problematic since 0/0 isn't defined in the first place. You can't get a single value as an answer (in other words ##\frac{0}{0}*\frac{0}{2}≠0##, it's an invalid operation).

Ok that makes sense. As fresh_42 said in a prior thread, to forget 0/0 as it's undefined and I was thinking about that when I came to 0/0 and knew it was undefined. My troubles were why the sin/sin=1 but if I used the limit laws for the numerator and denominator separately, it gave 0/0 which I did not want as it was undefined. So I starting wondering if my use of the limit laws was improper.
 
  • #6
opus said:
Isn't the purpose of the Limit Laws to be able to take bigger functions or expressions, and break them down into more manageable things i.e factoring things out?
You are missing the point. You are only allowed to do this if the individual limits are well defined. It is just wrong to factor out the limit of 1/x.

opus said:
Ok that makes sense. As fresh_42 said in a prior thread, to forget 0/0 as it's undefined and I was thinking about that when I came to 0/0 and knew it was undefined. My troubles were why the sin/sin=1 but if I used the limit laws for the numerator and denominator separately, it gave 0/0 which I did not want as it was undefined. So I starting wondering if my use of the limit laws was improper.
Very much so.
 
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  • #7
Orodruin said:
You are missing the point. You are only allowed to do this if the individual limits are well defined. It is just wrong to factor out the limit of 1/x.
Ok so a two part question:

1) In my text, it states that "At each application of a limit law, the new limits must exist for the limit law to be applied". I didn't know exactly what this was referring to but now I think that it is applying to this exact case. Is this true?

2) In regards to your statement of "factoring out the limit of 1/x":

What about multiplying by 1/x? As an example, here is a problem we did in class.

Evaluate ##\lim_{x \rightarrow +∞}{\frac{3x-5}{8x+10}}##

Multiply the numerator and denominator by 1/x to get ##\frac{\left(\frac{1}{x}\right)3x-5\left(\frac{1}{x}\right)}{\left(\frac{1}{x}\right)8x+\left(\frac{1}{x}\right)10}##
$$=\frac{3-\frac{5}{x}}{8+\frac{10}{x}}$$
From here, we can see that as x approaches infinity, the secondary terms in the numerator and denominator go to zero, and the limit of the whole is ##\frac{3}{8}##
Similarly done for x approaching negative infinity.
 
  • #8
opus said:
2) In regards to your statement of "factoring out the limit of 1/x":

What about multiplying by 1/x?

I think Orodruin means ##lim_{x→0}\frac{1}{x}##. This is undefined and certainly not equal to zero. To see why, take the individual limits from the positive and negative directions:

##lim_{x→^-0} \frac{1}{x} = \frac{1}{-0}=-∞##
##lim_{x→^+0} \frac{1}{x} = \frac{1}{0} = +∞##

Edit: the expressions ##\frac{1}{-0}## and ##\frac{1}{0}## are not defined and should not be used. The limit equations above work just fine by omitting them.

Since the two limits are not the same the limit does not exist.

Multiplying by ##\frac{1}{x}## gives:
##lim_{x→^-0} \frac{1}{x} =\frac{\frac{1}{x}1}{\frac{1}{x}x} =\frac{\frac{1}{x}}{1}=lim_{x→^-0}\frac{1}{x}=-∞##
##lim_{x→^+0} \frac{1}{x} =\frac{\frac{1}{x}1}{\frac{1}{x}x} =\frac{\frac{1}{x}}{1}=lim_{x→^+0}\frac{1}{x}=∞##
This is, of course, as it should be, since we only multiplied by 1.
 
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  • #9
opus said:
1) In my text, it states that "At each application of a limit law, the new limits must exist for the limit law to be applied". I didn't know exactly what this was referring to but now I think that it is applying to this exact case. Is this true?
An example of what your book is saying is this:

##\lim_{x \to a}f(x)\cdot g(x) = \lim_{x \to a}f(x) \cdot \lim_{x \to a}g(x)## provided that both limits on the right exist. If either limit on the right doesn't exist, then the operation isn't valid.
 
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  • #10
Drakkith said:
To see why, take the individual limits from the positive and negative directions:
##lim_{x→^-0} \frac{1}{x} = \frac{1}{-0}=-∞##
##lim_{x→^+0} \frac{1}{x} = \frac{1}{0} = +∞##
We don't even write either ##\frac 1 {-0}## or ##\frac 1 0##, since neither of these expressions is defined.

Instead, these would be written as
##\lim_{x→^-0} \frac{1}{x} = -∞##
##\lim_{x→^+0} \frac{1}{x} = +∞##
 
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  • #11
Mark44 said:
We don't even write either ##\frac 1 {-0}## or ##\frac 1 0##, since neither of these expressions is defined.

Instead, these would be written as
##\lim_{x→^-0} \frac{1}{x} = -∞##
##\lim_{x→^+0} \frac{1}{x} = +∞##

My mistake. It's been a while since I've worked with limits. Please let me know if anything else in my posts are incorrect.
 
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  • #12
opus said:
2) In regards to your statement of "factoring out the limit of 1/x":

What about multiplying by 1/x? As an example, here is a problem we did in class.

Evaluate ##\lim_{x \rightarrow +∞}{\frac{3x-5}{8x+10}}##

Multiply the numerator and denominator by 1/x to get ##\frac{\left(\frac{1}{x}\right)3x-5\left(\frac{1}{x}\right)}{\left(\frac{1}{x}\right)8x+\left(\frac{1}{x}\right)10}##
$$=\frac{3-\frac{5}{x}}{8+\frac{10}{x}}$$
There is an easier way:
##\frac{3x-5}{8x + 10} = \frac{x(3 - 5/x)}{x(8 + 10/x)}##
Now, ##\lim_{x \to \infty}\frac{x(3 - 5/x)}{x(8 + 10/x)} = \lim_{x \to \infty} \frac x x \cdot \lim_{x \to \infty}\frac{3 - 5/x}{8 + 10/x}##
The first limit clearly (??) exists, since for any finite value of x, x/x equals 1. The second limit is straightforward, as the fractional terms in the numerator and denominator both go to zero.
opus said:
From here, we can see that as x approaches infinity, the secondary terms in the numerator and denominator go to zero, and the limit of the whole is ##\frac{3}{8}##
Similarly done for x approaching negative infinity.
 
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  • #13
Drakkith said:
Please let me know if anything else in my posts are incorrect.
No worries, Drak! :oldbiggrin:
 
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  • #14
Mark44 said:
The first limit clearly (??) exists, since for any finite value of x, x/x equals 1.

Should that be non-zero value of x? Or perhaps non-zero, finite value of x?
 
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  • #15
opus said:
Isn't the purpose of the Limit Laws to be able to take bigger functions or expressions, and break them down into more manageable things i.e factoring things out? And I'm not sure what you mean by "converges". I haven't heard that term yet. Does it mean that the limit is the same as the x value gets approached from both sides?
Saying a limit converges means the limit exists. If you say the limit of f(x) as ##x \to a## is equal to ##L##, you mean that the function f(x) approaches, or converges to, ##L## from both sides as ##x \to a##.

The property of limits you're using is
$$\lim_{x \to x_0} f(x)g(x) = \left[\lim_{x \to x_0} f(x)\right]\left[ \lim_{x \to x_0} g(x) \right],$$ and the equality only makes sense if all three limits exist. If one or more of the limits didn't exist, then one or both sides of the equation would be undefined. For example, it would be incorrect to say
$$\lim_{x \to 0} \frac xx = \left(\lim_{x \to 0} x \right) \left( \lim_{x \to 0} \frac 1x \right)$$ because the last limit doesn't exist.

So if you're going to use that property, you need to make sure the individual limits in the product exist. In your problem, it's not wrong to split the limit up the way you did as all three limits converge, but it's pointless. The problem you were trying to work around was avoiding the indeterminate form 0/0 that you got from plugging 0 into the original expression. By splitting up the limit after the manipulations you did, you ended up with another limit of the form 0/0, which is exactly the problem you started with. Instead, you could have just said
\begin{align*}
\lim_{\theta \to 0} \frac{1-\cos \theta}{\sin \theta} &= \lim_{\theta \to 0} \frac{1-\cos \theta}{\sin \theta} \\
&= \lim_{\theta \to 0} \frac{\sin^2 \theta}{\sin \theta\,(1+\cos\theta)} \\
&= \lim_{\theta \to 0} \frac{\sin \theta}{1+\cos\theta}.
\end{align*} The last limit you could evaluate easily by setting ##\theta## to 0.
 
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  • #16
Drakkith said:
Should that be non-zero value of x? Or perhaps non-zero, finite value of x?
Yes, I should have stipulated that x isn't zero, but for the limit at hand, with x approaching infinity, we can assume that we're not talking about a value of zero for x.
 
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  • #17
Mark44 said:
Yes, I should have stipulated that x isn't zero, but for the limit at hand, with x approaching infinity, we can assume that we're not talking about a value of zero for x.

My mistake, I actually didn't notice that the limit was going to infinity, not to zero. Now your statement makes perfect sense.
 
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  • #18
Thanks guys I think I'm starting to get the hang of it more now. By "if the limits exist" i took that as, if you're limits exist with what you have right now, then you can use the limit laws. I bought myself Schaums 3000 Calculus problems book and I'm just going to hammer these types of problems.
 
  • #19
vela said:
Saying a limit converges means the limit exists. If you say the limit of f(x) as ##x \to a## is equal to ##L##, you mean that the function f(x) approaches, or converges to, ##L## from both sides as ##x \to a##.

The property of limits you're using is
$$\lim_{x \to x_0} f(x)g(x) = \left[\lim_{x \to x_0} f(x)\right]\left[ \lim_{x \to x_0} g(x) \right],$$ and the equality only makes sense if all three limits exist. If one or more of the limits didn't exist, then one or both sides of the equation would be undefined. For example, it would be incorrect to say
$$\lim_{x \to 0} \frac xx = \left(\lim_{x \to 0} x \right) \left( \lim_{x \to 0} \frac 1x \right)$$ because the last limit doesn't exist.

So if you're going to use that property, you need to make sure the individual limits in the product exist. In your problem, it's not wrong to split the limit up the way you did as all three limits converge, but it's pointless. The problem you were trying to work around was avoiding the indeterminate form 0/0 that you got from plugging 0 into the original expression. By splitting up the limit after the manipulations you did, you ended up with another limit of the form 0/0, which is exactly the problem you started with. Instead, you could have just said
\begin{align*}
\lim_{\theta \to 0} \frac{1-\cos \theta}{\sin \theta} &= \lim_{\theta \to 0} \frac{1-\cos \theta}{\sin \theta} \\
&= \lim_{\theta \to 0} \frac{\sin^2 \theta}{\sin \theta\,(1+\cos\theta)} \\
&= \lim_{\theta \to 0} \frac{\sin \theta}{1+\cos\theta}.
\end{align*} The last limit you could evaluate easily by setting ##\theta## to 0.

Yeah I did feel like I ended back to where I started with that problem! One thing I need to get used to is the idea that there is more than one way to solve a problem not just a single right way and I need to practice to find out what works and what doesn't so much.
 

Related to Why Does Splitting Limits Cause Problems with 0/0 Forms?

1. What is the meaning of "0/0" in limit laws?

The expression "0/0" is known as an indeterminate form in mathematics. It does not have a specific numerical value and cannot be solved using basic algebraic methods.

2. Can the limit of a function at a point where the denominator is 0 be calculated?

No, the limit of a function at a point where the denominator is 0 cannot be calculated. This is because the expression becomes undefined and does not follow the rules of limit laws.

3. What are some strategies for evaluating limits involving 0/0?

One strategy is to factor the numerator and denominator and see if any common factors can be canceled out. Another strategy is to use L'Hospital's rule, which states that the limit of a fraction can be calculated by taking the derivative of the numerator and denominator and then evaluating the limit again.

4. Can the limit of a function be 0/0 at a point where the function is defined?

Yes, the limit of a function can be 0/0 at a point where the function is defined. This does not violate the limit laws as long as the function is continuous at that point.

5. How do limit laws apply to functions that have removable discontinuities?

If a function has a removable discontinuity, then the limit laws can still be applied as long as the discontinuity is removed. This means that the function is redefined at the discontinuity point so that it becomes continuous, and the limit can then be evaluated using the new definition.

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