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# L'Hôpital's rule: limit at 0 example

Sal uses L'Hôpital's rule to find the limit at 0 of (2sin(x)-sin(2x))/(x-sin(x)). Created by Sal Khan.

## Want to join the conversation?

• How do you know when to stop applying the rule?
• L'Hopital's rule only applies when the expression is indeterminate, i.e. 0/0 or (+/-infinity)/(+/-infinity). So stop applying the rule when you have a determinable form. In the case of the trig functions, they repeat themselves on the 4th der, so if, after the third application of L'Hopital's rule to the given expression we had gotten undefined, then we would know that the limit did not exist.
• How do you know when the limit doesn't exist at all?
• quoting rhorcher: "I think it's worth note that L'H Rule does not apply to all undefined forms just some."
Well, when you take the limit and arrive at an answer of 0/0, this is actually an INDETERMINANT. An example of an UNDEFINED number would be 1/0 or infinity.
So what I THINK is that L'Hospital's rule may not apply to limits that are UNDEFINED.
• Could someone explain what he did at ? I get that he took the derivative, and that the derivative of cosx=-sinx, but how did he get the 4 in front?
• you find the derivative of cos(2x) with the chain rule : it's the product of the derivative of the intern function by the derivative of the extern function :
d/dx[cos(2x)] = d/dx[2x]d/dx[cos](2x)
= 2 * -sin(2x)
So, d/dx[-2cos(2x)] is -2 * d/dx[cos(2x) = -2*2
-sin(2x) = 4sin(2x)
• When do you give up...?
• Never give up! Always strive to keep learning. I didn't understand a topic in this class last week, but I returned a few days later, rewatched all the videos, did the problems slowly, and was finally able to understand it. If you feel stuck, encourage yourself to keep going and you will eventually understand the material.
• Could someone please help me understand from to better. If we used the product rule, then (sin)(2x) will be cos(2x)+ 2sin. I am really lost.
• You may be accustomed to your instructor using more parentheses. Note that the function in question is meant to be sin(2x), NOT sin(x) (2x). This function is a composition function (double x, then apply sin). As such, you should use the chain rule, not the product rule.
• why wasn't the quotient rule used to to find the derivative at .
• because we don't want the derivative of the quotient but the quotient of the derivatives
• Isn't using L'Hopital's rule the easiest way to find out if a limit exists? Why don't we use it all the time, instead of using algebraic methods? Is it because you always need two functions (one for the numerator, one for the denominator)?
• l'Hopital's is true ONLY if you have a 0//0 or an ∞ / ∞ form. If you have any other form, it is not true.
For example:
lim x→ 5 {{x² / (x²-20)} = 5

But if you try to do it with l'Hopital's you get:
lim x→ 5 {{2x / (2x)} = 1, which is wrong.
• In looking at some of the discussion, one of the posters said that you stop applying L'Hopital's rule when "the answer you get isn't 0/0 or infinity over infinity". What if I get infinity over some number or some number over infinity? Would I continue, or would this reflect a mistake?
• When you have a limit that is a number over infinity it's equal to 0 and when you have a limit that is infinity over a number it's equal to infinity
• He had to differentiate 3 times to get an answer, but say we have a different function. Are there functions for which you can differentiate n times without it getting an answer? If not, how many times should you try differentiating the top and bottom before giving up? Thanks in advance
• Some limits remain indeterminate no matter how many times you apply l'Hôpital's rule. Some will eventually work out, but you have to apply l'Hôpital's rule many, many times.

So, you just need to observe whether you are making progress. You might look, for example, whether there is an exponent that is changing with each differentiation that is getting closer to something you can take the limit of?

So, no, there is not a set number of times you should apply l'Hôpital's rule. But, if it doesn't work within a reasonable amount of time, you might want to look at other means of finding the limit.