The answer is correct but I made a mistake when I did L'Hopital's Rule for the 2nd time. See pinned*

This calculus tutorial will help you understand how to evaluate the limit of 1/sin(x)-1/x as x goes to 0 by using L'Hopital's rule. This is a typical calculus 1 or calculus 2 limit and a common exam problem. It is in the form of inf-inf but is it 0?

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[nobodyspecial]

You made a mistake I believe with the second use of L'hospital's rule. You differentiated xcosx to x*-sinx + sinx * 1 when it should be x*-sinx + cosx * 1. So the final answer should be 0/2

[gogl]

3:24 For the second usage of L'Hopital shouldn't the denominator be -x(sin x) + 2 cos x instead of -x(sin x) + sin x + cos x?

[spiderjerusalem]

For the sake of brevity, let lim f(x) denote the limit of f(x) as x→0.

lim(csc x – 1/x)
= lim ((x–sinx)/x³)(x/sinx)x.

Since the limit of each of these three terms exists, by algebraic limit theorem,


= ⅙(1)(0) = 0

[arseniix]

Applying the engineering trick that sin(x) = x for small x, we can just do 1/x - 1/x = 0 and deal with it 😂

Chat, where I'm wrong?

[billcook]

This may be sloppy, but we know lim x-> 0 sin(x)/x = 1. So the two terms in the original are essentially equal near 0. So their difference is essentially 0 near 0.

[umylten]

I see many comments mention the use of sin(x) ≈ x. Based on the video it intuitively cannot work because he used L'Hospital *twice* i.e. he differentiated twice, while sin(x) ~ x is a 1st order approximation (this is only an intuition, I never said it's a proof lol).

However if we consider the next term in the Taylor series, sin(x) ≈ x - x³/6. Thus we get

x/sin(x) ≈ x/(x - x³/6) = 1/(1- x²/6) ≈ 1 + x²/6

(In the last equality, use 1/(1-u) ≈ 1+ u)

In the end we get

1/sin(x) - 1/x ≈ 1/x + x/6 - 1/x = x/6 -> 0.

[GreenMeansGOF]

After the first L’Hopital’s rule, we have



which has a limit if 0/(1+1) which is 0.

[meurdesoifphilippe]

Let f and g be 2 functions such that f(x) ~g(x) when x->0. Does it means that 1/f - 1/g - >0 when x->0? No. Here it does but only by chance. Set f(x) =x and g(x) =x+x^p for some p>1. f~g as x->0, ok. 1/f-1/g = 1/x * (1-1/(1+x^{p-1})). But 1/(1+u) ~ 1-u when u->0 then 1/f-1/g ~ x^{p-2} (don't forget the 1/x) which goes to 0 if p>2, 1 if p=2, and even infty for 1<p<2!!! Here sinx ~x-x^3/6 so it's the same as p=3 (with a constant too). The equivalence of x and sin x allows you to substitute but only in multiplications and division,

[JeeAsp]

You can do it without L'Hopital too ( Just for extra )
multiply divide by x+sinx in Nr and Dr
then make the equation as x^2 -sin^2x / xsinx ( x+Sinx)
rewrite it as x^2 - ( 1+cosx)(1-cosx)/xsinx(x+sinx)
Now take x^2 common from the Nr
and x from xsinx and another x from x+sinx

You'll get it done

[JohnSmith-eglc]

Solution by the use of the squeeze theorem:
We have lim x-> 0 of (1/sin(x) - 1/x) = lim x-> 0 of (x-sin(x))/(xsin(x)). We know that cos(x) < sin(x)/x < 1. Therefore (x-sin(x))/(xsin(x)) = (x-xsin(x)/x)/(x^2sin(x)/x) = (1 - sin(x)/x)/(xsin(x)/x) > (1 - 1)/(x*1) = 0. Similarly (x-sin(x))/(xsin(x)) = (1 - sin(x)/x)/(xsin(x)/x) < (1 - cos(x))/(xcos(x)) = (-1/cos(x)) * ((cos(x) - cos(0))/x). Now, the limits lim x-> 0 (-1/cos(x)) and lim x-> 0 (cos(x) - cos(0))/x both exist and they are finite (the latter is just the derivative of cos(x) evaluated at x = 0). We therefore have lim x-> 0 (-1/cos(x)) * ((cos(x) - cos(0))/x) = [lim x-> 0 (-1/cos(x))] * [lim x-> 0 (cos(x) - cos(0))/x] = -1 * (-sin(0)) = 0. It then follows from the squeeze theorem that lim x-> 0 (1/sin(x) - 1/x) = 0.

[Metaverse-df]

Maclaurin is powerful when the variable is approaching 0

[Metaverse-df]

0:49 It's far faster to use Maclaurin.

[isavenewspapers]

Honestly would've been kinda funny if 1 / sin(x) had been written as csc(x).

[davidbrisbane]

We could use limit sinx/x = limit x/sinx = 1, as x goes to zero to solve this problem.

[Rishith]

can we solve by using the approximation that : when lim x -->0 : sinx ~ x ??
as this would give us the answer instantly....

[MyNordlys]

You made a BIG mistake with (xcosx)'

[Negan_]

Use L.H rule direct in original funtion...and we get zero immediately🎉

[MrMasterGamer]

lim_x->0 (1/sinx - 1/x) =
lim (1/sinx - sinx/xsinx) =
lim(1/sinx) - lim(sinx/x)lim(1/sinx) =
lim(1/sinx) - lim(1/sinx) =
0

[okaro]

Could one use the fact that lim x->0 sin x / x = 1 for this? Intuitively it is clear that it causes a zero but how would one do it rigorously?

[xinpingdonohoe]

Usefully, sin(x) behaves like x at 0. How would we say this? Perhaps sin(x)∈Θ(x)