Question: Can you use the special trig limit to find lim as x approaches 0 of x sin(1/x)? It looks like it could be rewritten as lim as x approaches 0 of (sin 1/x)/(1/x) to get 1. However when I use the Squeeze Thm to find its limit, I get zero. Why???
The special trig limit is "limit as x goes to 0 of (sin x)/x". If you want to generalize, it's "limit as STUFF goes to 0 of (sin STUFF) / STUFF". But "limit as x goes to 0 of (sin 1/x) / (1/x)" doesn't match this general form because all three occurrences of "STUFF" must be the same (or at least constant multiples of each other). If you want to apply the special trig limit to (sin 1/x) / (1/x), then 1/x has to go to zero, which means x itself goes to positive or negative infinity.
Like I said starting around 4:10, the point of the substitution (and substitutions in general) is just to take something and make it look less complicated and/or more familiar. Replacing 2x with t makes the expression look identical to the first limit in the fun fact in the beginning of the video (except that the variable there was x and not t but that's completely irrelevant), and the value of that limit is 1. The limit you're asking about is 2 because of the factor of 2 being multiplied by the limit.
sinx/x and x/sinx aren't equal. The limit as x approaches 0 of sinx / x is equal to the limit as x approaches 0 of x/sin x. (I know that's what you meant but other people reading this might not know, plus it's an important distinction that can result in lost points if not made properly.) The reason that the limits are equal is because the limit as x approaches c of f(x) is equal to 1/(the limit as x approaches c of [1/f(x)]), as long as the limit as x approaches c of f(x) is not zero. In this case we have f(x) = sinx/x, so then 1/f(x) = x/sinx, and we have c=0. So the equality tells us that limit as x approaches 0 of sinx/x = 1/(limit as x approaches 0 of x/sinx). Since the limit as x approaches 0 of sinx/x is 1, then the equality becomes 1 = 1/(limit as x approaches 0 of x/sinx), which necessary means the limit as x approaches 0 of x/sinx = 1.
I didn't say (cos x - 1)/x and (1 - cos x)/x are the same. Their LIMITS are the same if we take the limit as x goes to 0. And yes, 1-cos x = -(cos x - 1), but since the limit is zero, we just get -0 for the other one. And -0 = 0 anyway, so the limits are the same.
Caleb Tshionza It's subjective as to what's easier. But using a double angle identity only works for double angles. The method in example 1 works regardless of the constant multiple on x in the sine function.
This Video might not be the best quality, but your the only person I've understood so far; thanks dude!
No problem, glad to help! This is one of my earlier videos from when I had a bad setup. My later videos have better quality.
Thanks for the detailed information about how to find the limit of this basic trigonometry functions....very helpful!
this really helped me because i wasn't able to start my course in calculus II. Thanks a lot! :D
Great job man, keep it up you're videos are getting me through calc!!
Thanx man You are an awsome teacher
Those are really FUN FACTS
Question: Can you use the special trig limit to find lim as x approaches 0 of x sin(1/x)?
It looks like it could be rewritten as lim as x approaches 0 of (sin 1/x)/(1/x) to get 1. However when I use the Squeeze Thm to find its limit, I get zero. Why???
The special trig limit is "limit as x goes to 0 of (sin x)/x". If you want to generalize, it's "limit as STUFF goes to 0 of (sin STUFF) / STUFF". But "limit as x goes to 0 of (sin 1/x) / (1/x)" doesn't match this general form because all three occurrences of "STUFF" must be the same (or at least constant multiples of each other). If you want to apply the special trig limit to (sin 1/x) / (1/x), then 1/x has to go to zero, which means x itself goes to positive or negative infinity.
I understood this, thanks!
why did you replace 2x with t? and how did that tell you that the limit = 2?
Like I said starting around 4:10, the point of the substitution (and substitutions in general) is just to take something and make it look less complicated and/or more familiar. Replacing 2x with t makes the expression look identical to the first limit in the fun fact in the beginning of the video (except that the variable there was x and not t but that's completely irrelevant), and the value of that limit is 1. The limit you're asking about is 2 because of the factor of 2 being multiplied by the limit.
ohh ok I get it now! thanks so much!
7 years later, I was just asking myself the same thing. I guess I have to get used to some calc techniques... it reminds me of identities
thank you man
i understood
جزاك الله خیرا
Thank you so much.
Thank you.
Thx alot
can i ask a question... i`m just curious please answer this
proof why sinx/x is equals to x/sinx T.T
sinx/x and x/sinx aren't equal.
The limit as x approaches 0 of sinx / x is equal to the limit as x approaches 0 of x/sin x.
(I know that's what you meant but other people reading this might not know, plus it's an important distinction that can result in lost points if not made properly.)
The reason that the limits are equal is because the limit as x approaches c of f(x) is equal to 1/(the limit as x approaches c of [1/f(x)]), as long as the limit as x approaches c of f(x) is not zero. In this case we have f(x) = sinx/x, so then 1/f(x) = x/sinx, and we have c=0. So the equality tells us that
limit as x approaches 0 of sinx/x = 1/(limit as x approaches 0 of x/sinx).
Since the limit as x approaches 0 of sinx/x is 1, then the equality becomes
1 = 1/(limit as x approaches 0 of x/sinx),
which necessary means the limit as x approaches 0 of x/sinx = 1.
i love you
(cosx-1)/x is not the same with the other one cuz u need to change signs to transform it like this 1-cosx = -(cosx-1) dont forget the minus sign!!
I didn't say (cos x - 1)/x and (1 - cos x)/x are the same. Their LIMITS are the same if we take the limit as x goes to 0. And yes, 1-cos x = -(cos x - 1), but since the limit is zero, we just get -0 for the other one. And -0 = 0 anyway, so the limits are the same.
unogona iwe une swag
Blurry video, but the math is clear.
Wouldn't it be easier to us double angle identity? ?
Caleb Tshionza It's subjective as to what's easier. But using a double angle identity only works for double angles. The method in example 1 works regardless of the constant multiple on x in the sine function.
Dude go for more complicated sums ....
What does this video have to do with sums?
This shit aint fun..
THANK YOU SO MUCH