Direct substitution of x in the original equation yields 0/0 so you can use L’Hopital’s rule.

That doesn't follow. By the same argument, the limit as x tends to zero of sin x / x wouldn't exist, but it is 1.

Just because you got the right answer doesn’t mean you did the correct work, in this case you evaluated with 0 and received an undefined value, but an undefined value doesn’t mean that a limit doesn’t exist. You would have to either simplify, or use a graph/table approaching from above and below to tell the answer

the whole point of learning limits of functions is so that you can determine limits without graphing...anyone could graph a function and pick where the points cluster around as the limit

Angea H. He’s writing with a mouse dumbass

Yet he thought it was funny to mock the Lohan.

Scott Collier how is your location relevant?

0/0 is indeterminate hence the need to use l'hopital's rule. There are instances where 0/0 is the result of substitution and yet we still have a definite limit.

I don't know calculus btw just a big mean girls fan :)

Mr. Wan probably paint on a pc

Try to refrain from telling people they are incorrect when you yourself dont even understand what youre talking about, as you can read in the article it states if a function experiences undamped oscillations, and its using an example as x approaches c which is pos/neg infinite in this case. The case in the video is not experiencing any undamped oscillations as x is approching a finite value which in this case is zero.

If you want to follow L’Hopital’s rule to the tee then you would take the limit from the left and right sides of zero and find that you get infinity and neg infinity which results in a divergent answer but doing it the way he did it is not wrong. Doing it the way he did it is breaking L’Hopital’s rule but not breaking the mathematical concept behind it

Just because received an undefined value from an attempted evaluation, it doesn’t necessarily indicate that the limit doesn’t exist. If you can’t simplify, or evaluate then you would have to use a table or a graph. Approaching 0 from both sides would yield a DNE but you should never say it does not exist just because your evaluation yielded an undefined value.

Mr. Who?

If you have an infinitely large number on the bottom, doesn't that give you 0 as the answer? And in this problem, don't you always get a 0 on the bottom so the limit DNE

+Taranjit Dogra Yep, you're there :) You're getting an infinitely small number on the denominator (since as x ---> 0, 1-cos^2(x) ---> 0), so the whole number is getting larger and larger! As x tends to 0, the equation tends to infinity :)

+William Stevens-Harris If you have to solve lim (x>0) f(x) / g(x) where g(0) = 0, that does not guarantee that the limit is tending towards positive or negative infinity. For example when you have to solve lim (x>0) x / x, you will probably be able to see that this should be equal to 1. But the denominator is equal to 0 when x = 0. The reason this can happen is because the numerator is also equal to 0 when x = 0. When both the numerator and the denominator equal zero when x = 0, you need to be more clever when you try to find the limit. In this case l'Hopital will be good enough to find out why the answer actually is that the limit does not exist.

+SmileyMPV yeah that's what did. I took the derivative of top and bottom and noticed that the bottom will always be 0 since sin(0) will always be 0.

+Enderman1323 Suppose we had the limit of Meeeooowwww

Enderman1323 yeees it was hahaha

She actually got -1

An ipad app called Explain Everything. Its a paid app but there are several similar apps which are free.

How rude.

On the streets he's known as "El Hospital," cuz that's where he put all his mathematical adversaries.

The one-sided limits are, but since it's positive infinity from one side and negative infinity from the other, for the two-sided (general) limit it's best to just say it doesn't exist.

+BlakeryMath I know how you can tell whether or not a limit exists if you were to give me a graph, but how can you tell whether the limit is negative infinity or does not exist if you have a value over 0 by direct substitution?

+Gary Lu It's more like "really close by" substitution. You can get an idea of what is happening if you plug in numbers close to zero on both sides (positive and negative) and see what happens to the y values. For instance, I can get some pretty good evidence that the lim x->0 5/x does not exist, even if I didn't know what the graph looked like, by realizing that 5/0.00001 is a very large value, and 5/-0.00001 is a very large negative value, so as x -> 0 from the right the function tends toward positive infinity, and as x -> 0 from the left the function tends toward negative infinity.

No, it has 4 conditions -limit mus be 0/0, inf/inf -g’(x) cannot go to zero -f(x) must be differentiatable everywhere in the neighborhood except Maybe at a -limit if it exists then they are equal Otherwise the conclusion is inconclusive and requires a different method

Incorrect - you can not use LH twice because the numerator is not zero. We were able to use LH the first time since the function was indeterminate (0/0) as x -> 0. After we use LH once it is now -2/0 which is NOT indeterminate form. Ergo the limit does not exist!

But then you'd be misusing the rule, since the second limit is not of indeterminate form.

I call it La Hospital's rule, too. Lmao.