Limit of (1-cos(x))/x as x approaches 0 (Proof) | Derivative rules | Science Valhalla

Limit of (1-cos(x))/x as x approaches 0 (Proof) | Derivative rules | Science Valhalla

Hi, welcome again to our YouTube channel Science Valhalla. In today's video, we'll see the limit of cos x-1/x as x approaches 0. We will see the geometrical proof. limit of cos x-1/x as x approaches 0 can be proved using the squeeze theorem as well. cos x-1/x is actually a very important theorem because we use it in the derivative derivation of Trigonometry functions. Always remember the angle is x here, sometimes we use theta instead of x, so don't get confused. And x is in radians. I hope you will enjoy the animated video.

The same type of problems also exist, you can look at them also:-
1. limit of sinx/x as x approaches 0
2. lim x → 0 sin 1/x
3. limit of cos(1/x) as x approaches 0
4. lim 1-cosx/x^2 as x approaches 0
5. limit of cos x as x approaches 0
6. limit of 1 cosxx2 as x approaches infinity
7. lim x- 0 (1-cosx)/x^2

Other searched terms:-
"limit of cos x as x approaches 0"
"limit of cos(1/x) as x approaches 0"
"lim x- 0 (1-cosx)/x^2"
"1-cos x/x squeeze theorem"
"cos x/x limit"
"1-cos x formula"

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[mamtaswajeetrai]

From childhood I never was convinced with any proofs without a visual representation…

Tons of thanks… you don’t know how much it helped me.

I m subscribing you, right now.🙏🙏

[marcoponts]

This is no proof and although the idea with visualizing it is nice, the solution is by no means something that would be accepted as an answer on any exam. I think it's cool that you do math videos! Don't get me wrong. But I think there is some improvement possible, especially if you write (proof) in the name of the video. By all means, this is no proof.

As mentioned in a cool comment before, you can use de l'Hôpitals rule. You could also write the Taylor expansion of cosine and cancel the x. Then you see that it goes to zero.

[matematicasmad]

This and the proof for sin(x)/x are tiene better videos on the subject I could find... thanx.

[holyshit]

It can be reduced to the sinx/x limit in two ways
1.
Multiply top and bottom to get sin(x)/x
2.
Use Pythagorean identity and cos od double angle formula
cos(x)=cos(x/2)^2-sin^2(x/2)
1 = cos(x/2)^2-sin^2(x/2)
cos(x) - 1 = -2sin^2(x/2)
-2sin^2(x/2)/x
-1/2x(sin(x/2)/(x/2))^2
sin(x)/x can be calculated using squeeze theorem
Necessary inequalities can be found by comparing areas of triangles and sector

[FridgeGames]

If I understand the last part of the video, you say that since (1-cosx) decreases way faster than x does, then the ratio approaches 0. This is basically a visual representation of L'Hopital rule. You check how fast the numerator is increasing/decreasing relative to the denominator by doing the derivative of both terms. For instance, logarithm (lnx) grows slower than linear (x+c), which grows slower than exponential (e^x), and so on. The term that grows more rapidly "wins" against the slower term in a sense.

[herlysqr]

Some people are saying "you can use L'Hopital rule" but no, you can't, because to use it you should know what is the derivative of cos(x), but to know that you should first know what is precisely the limit of this video.
So using L'Hopital rule would be circular reasoning.

[subhadeepmondal]

cosx-1=-2sin^2(x/2)
now lim x tends to 0 (cosx-1)/x=lim x tends to 0 (-2sin^2(x/2))/x now you can do this from here.
isn't this is easier than that
or you can do L'Hospitals that is far more easy to visualize and to do.

[AbouTaim-Lille]

Use the Taylor expansion in a neighborhood of zero and it is over

[SKAOG]

Can't you just approximate cos x = 1-(x^2)/2, do some algebraic manipulation and show that whats left is -x/2, which tends to 0 as x tends to 0?

[desertrainfrog]

Interestingly enough, the limit as x-->0 of [1-cos(x)]/x also equals 0.

[klementhajrullaj]

But, if will was it (cosx-1)/x^2, the limit will wasn't 0, but will was -1.

[Raots-wvhh]

It can be much easily solved by simplifying numerator and then apply limits.
The last part of your explanation little confusing.

[nmgactor]

Or, since when 0 is plugged in you end up with the the indeterminate form 0/0, you can use L’Hospital’s Rule. Once you take the derivative of both sides you get -sin(x) or sin(x). Which when evaluated at 0 gives you 0.

Another option is to multiply by the conjugate pair. Let’s do cos(x)-1/x.
Cos(x)-1/x * cos(x)+1/cos(x)+1
Cos^2(x)-1/x(cos(x)+1) Pythagorean Identity
-sin^2(x)/x(cos(x)+1
Sin(x)/x * -sin(x)/cos(x)+1 Special Limit
1*-sin(x)/cos(x)+1
-sin(0)/cos(0)+1
0/1+1
0/2
=0

Phew done, and it’s very similar work with 1-cos(x)/x so I won’t do that one. Those are both mathematical proofs for it that are more algebraic compared to the more visual geometric proof, which is still very cool. Goes to show that there’s a lot of different ways to come to the right answer in math.

[yiquny]

Not convincing. This will mislead students.