tricky calculus limits you need to be careful!

We will discuss the limit of sin(ln(x))/x x goes to 0+ vs. the limit of sin(ln(x))/x as x goes to inf. These two limits are suitable for precalculus and calculus 1 students.

[lucasfrykman]

"This is badly behaved" yeah tell that limit who's boss.

[TheAutoBeef]

the first function converges to 0, the second one converges to 1 ((sin 1/x)/(1/x) the same as (sin y)/y when y goes to 0)

[emanueleusai]

For the 1st one the answer is 0, lim x->o+ (x*sin(1/x), let t=1/x, therefore you have lim t->inf (1/t)*sin(t)=0
The 2nd one: lim x->inf (x*sin(1/x), let t=1/x therefore you get lim t->0+ (1/t)* sin(t)= lim t-> 0+ (sin (t)/t) then use l'hopital's rule and you will get lim t->0+ (cos(t))=1

[omarifady]

For the limit question
As x approaches 0, we just pluging
We get 0*sin(1/0)
=0*sin(inf)=0*(number between-1 and1) = 0
And as x approaches inf
We get
Inf*sin(0)=inf*0
We bring x down down and use l’hopital rule
We get cos(1/x)=cos(1/inf)=cos(0)=1

By the way, the graph of this function is very cool!

[gplor]

Your accent is so clear and peaceful for some reason and it makes things easier to understand.

[mattiarigo]

1) as x goes to 0+ sin (1/x) will be between -1 and 1, but in anyway if it's multiplied by x it will become 0
2) as x goes to infinity sin (1/x) approaches 0, so you have a very very big number times a very very small number which equals 1

I hope it's somewhat right, correct me if I'm wrong

[letsgetiton]

wow this is this is actually asked in a comment section at your previous vid and u actually solve it!! i think that guy who asked this would be happy

[stephomn]

clearly the first one is undefined and squeeze theorem for the second one which results to 0

[bhangbhosda]

First one answer to the question you asked at the end of the video will be zero and 2nd one will also be zero. Is it correct?

[MiguelFernandez-ixmi]

My guess:
1) +/- 1/infinity, approaches 0
2) approaches 1

[ttc]

Hello sir.can u help me
A= Limx->0 ((x^2-2x+2)e^x-2 )/x^3
B=Lim x->3 (x! - 6)/x-3
Don't use Loh'spital theorem

[alonbenjo]

In the questions just sub t=1/x then it is just like the lim you did in the video

[arulkapoor]

For 1 you have 0*sin(infinity), 0 multiplied by number between -1 and 1 is 0, so limit is 0. For 2, let u=1/x, lim u->0 sin(u)/u = 1

[polyreza]

Could you please solve for this limit:
Limit x^x^x^... when x ---> 0+

[holyshit]

Squeeze theorem for the second (limit to infinity)
Limits in question can be calculated in the same way as shown on video
and are good examples for practise

[clubstepdj]

lim(x->0+) [x sin(1/x)]
Make the x go down down
= lim(x->0+) [sin(1/x)/(1/x)]
On the top we have sin(1/0+) which is sin(inf)
And on the bottom we have 1/0+ which is inf
Like in example 2 of the video, the limit approaches 0


And when you approach inf instead, you will have sin(0)/0 which is a 0/0 situation so use L'H you get lim(x->inf) [{cos(1/x) (-1/x²)}/(-1/x²), cancel out the (-1/x²)s and you'll have cos(1/x), plugging x=>inf you'll get cos(0) which is 1

[mcmage]

For the first, it does not exist, since the function sin(ln(x)) as x goes to 0+ will be still on interval [-1;1] assuming the input is not complex.

For the second limit, its 0 since n/infinity -> 0

Now for questions
We can sub in v=1/x
And limits become:
1) Limit as v goes to infinity of sin(v)/v = 0 (if i remember correctly)
2) Limit as v goes to 0 of sin(v)/v = 1 (by sandwich theorem)

[purushotamgarg]

At 6:10, The first one is zero and the second is 1. Am I Correct???

[snbeast]

1. 0, because sine is always finite, and 0*finite is 0
2. 1, as infinity times zero is an indeterminate form, and doing some work (making the * x into a / 1/x, then doing L'Hopital's Rule) will get us cos(1/x)

[marche]

1st one. Sin(1/x) oscillates between -1 and 1. So by the squeeze theorem -1<sin(1/x)<1 ->
-x<xsin(1/x)<x. And the limit at 0+ is therefore 0.
2nd one.
Write as sin(1/x)/1/x. Use L'hospitals rule since its a 0/0 situation. Reduces to cos(1/x) which at infinity is cos(0)=1