This is a very famous limit

How do you do it 🤔 I had a similar question on my test just a few days ago and being new to limits having not even learned limits approaching infinity until yesterday, I was confused how to do it, everyone guessed including me, but I did attempt the question. The question I believe was limx->0(sin(x)/x).

so its the beard that makes him talk

bro messed up so hard, idk how he managed to not freak out

Nice. Wait. WTF? I thought I was confused before.

Normally my physics lesson when it gets so difficult like this during my 3rd year uni… i just answer either 0 or 1

LoL. He is Funny :)

L hospital invented in 1850
People in 1849 😂😂

You could use geometry to replace L'Hopital.

If he puts theta's value 0 , then it doesn't mean anything

Windows crashed

By sinx expression it is a single line problem😅

WHY WHY WHY AM I SEEING THIS AFTER MY CALC EXAM

/lim/(/theta)->0 sin(/theta)/(/theta). = 1

😂

Manjulika function 🥺

What you talking about😭

Standard limit 😔

Ans is 1

I used l'hopital's and got 1

Answer is 1

1

Finally he explaining 😂

At last you end up with sinx upon x with x tends to 0 which is equal to 1

It should actually be the limit as theta→0 from the right side only. If you have it as theta→0 you're actually evaluating the original limit for both x→-∞ and x→∞ and hoping they're equal.

Y->1/X so it becomes sin(Y)/Y when Y->0, by l'Hôpital it is clearly 1 and don't really care about other methods (don't even know if there is in that case)

Just saying, te answer would be 1, even if the limit was x going to negative infinity. I gave a question around this concept in an exam once, and let's just say... don't do it... a third of them deferentiated the top and bottom... and i had to give full marks for this because i forgot to cover it in class.... don't be me lol.

So does the limit approach infinity?

All the engineers laughing their ass off.

My engineering brain just go: sin(1/x) becomes 1/x so the limit goes to 1

=1

1

No offense to this man, but the way he tried to evaluate this limit is unnecessary complicated. When I saw the thumbnail for this video on my feed, I tried to evaluate it myself before watching. My first instinct was to use the limit law for the limit of a product (which meant breaking it up into the limit as x approaches infinity of x multiplied by the limit as x approaches infinity of sin(1/x). The first limit is infinity, and the second limit is zero because the denominator grows as x gets bigger, yielding inputs for the sine function that are closer and closer to zero. The sine of zero is zero, so the second limit is zero. Infinity multiplied by zero gives you zero. Therefore, the limit as x approaches infinity of xsin(1/x) is zero. Graph it on Desmos if you don’t believe me.

Fun fact due to the Taylor expansion of sin x*sin(n/x) as x goes to infinity is always equal to n

This is not wrong but very clever. Only real mathematicians know this trick😊

Sin thita by thita is 1

1

When theta approaches zero, sin theta also approaches zero, so that limits will give 1.

Use squeeze theorem !!!😅😂

it’s 1 bc of the fundamental trig limit

Isn't this the sinc(x) function which has a limit of 1 as x approaches 0?

I solved this you just use the lobital and step by step the solution is 1 because the deverative of sin is cos and cos(0)=1

it's simple but my eye tell me I can't do it

Bro was struggling at the end 😭 so in essence it’s 0

Since theta tends to 0, can't we use the small angle approximation? Sin theta~theta, so sin theta/theta tends to 1, which is the limit of the function.

IDIOT.

But isn't I/X=0 cuz 1/♾️ = 0 ?

And that's why (sin) (1/ infinite) =(sin) (0) = 0

Isn't that last part just solved by lhopitals rule? Its literally f(x)/g(x). I dont get what i am missing here...

Bro, L Hopitals rule😂😂😂😅😅

It's an inderminant case if we Directly put limit then it gives us 0 * (a number between -1,1) which is domain of sin theta hence 0

You can prove the hard way (squeeze theorem) or the easy way (L’Hôpital’s rule)