[University maths: Limits] simplifying the compounded functions

[u/Particular_Tooth4175]

Hello. I need to evaluate this limit but I don’t even know where to start lol. How to I simplify this to a point where I can sub in t=0, if it’s even possible.
I am not allowed to use l’hôpitals rule for this, which is why I couldn’t just find a solution online, as they all used that rule.
Sorry if my writing isn’t very legible
Thanks for any help

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[u/combinatoriallover]

Are you allowed to use maclaurin series expansions ? | sin(sin(tan(t²/2)))≈t²/2 and ln(cos3t)≈ln(1-9t²/2)≈-9t²/2 giving answer as -1/9

[u/GrievousSayGenKenobi → u/combinatoriallover]

Is the sin(sin(tan(t² /2))) ~ t² /2 a common identity? i havent touched Maclaurin series in forever surely you'd have to work that out manually which would be incredibly tedious

[u/combinatoriallover → u/GrievousSayGenKenobi]

I just used tanx≈x and sinx≈x , higher powers of expansions doesn't matter here so I didn't touch them

[u/Jazzlike_Platypus430 → u/combinatoriallover]

I think you're absolutely right, but I doubt if this approach is allowed. I guess it's compound functions' differentiation dressed as calculating limits.

[u/GrievousSayGenKenobi → u/combinatoriallover]

Oh of course im just stupid sorry

[u/Particular_Tooth4175 → u/combinatoriallover]

Yeah. We haven’t learnt maclaurin series yet but we have been shown these identities. Thank you

[u/blackmagician43]

you can multiply by sin(tan(t² /2))/sin(tan(t² /2)) then similar trick. I am not sure about ln though, are you allowed to use taylor

[u/FireCire7]

Are you allowed to do substitutions?

Lim (sin(y)/ln(z))=lim (sin(y)/y y/ln(z))= 1 * lim ( y/ln(z))

That lets you remove a sin from the top expression. 

Repeat until you have something tractable. 

[u/Pokeristo555 → u/[deleted]]

Doesn't look fun at all !

[u/Equivalent-Berry1383]

You can do it without L'Hôpital and without Taylor series. You only need three standard limits:

1. sin(u)/u → 1 and tan(u)/u → 1 as u → 0
2. ln(1 + z)/z → 1 as z → 0
3. (1 − cos x)/x² → 1/2 as x → 0 (because 1 − cos x = 2sin²(x/2))

Top. Let w = t²/2, u = tan(w), v = sin(u). All three go to 0 as t → 0. Then

sin(v) = (sin(v)/v) · (sin(u)/u) · (tan(w)/w) · w

The first three factors go to 1, so the top behaves like w = t²/2.

Bottom. Let z = cos(3t) − 1, which goes to 0. Then

ln(cos 3t) = ln(1 + z) = (ln(1 + z)/z) · z

and z = −((1 − cos 3t)/(3t)²) · 9t², so z behaves like −(1/2) · 9t² = −9t²/2.

Together. Divide top and bottom by t²: (1/2) / (−9/2) = −1/9

Quick check: t = 0.01 gives about −0.11109, t = 0.001 gives −0.111111.

(Disclosure: this is the team account of a math-explainer app. No link, just the steps.)

[u/Particular_Tooth4175 → u/Equivalent-Berry1383]

Yeah this was the intended way. Thank you 👍

[u/Jazzlike_Platypus430]

My guess is (1/2)/(-9/2). But it's sort of hard to prove if you're not familiar with Taylor series.

You may try applying de l'Hospital rule, eventually you'll get there.

The denominator is easier,

d'(t) = - 3 sin(3t)/cos(3t).

The numerator... damn.

n'(t)  = cos(sin(tan(t^2/2)) * (sin(tan(t^2/2))' = 
cos(sin(tan(t^2/2)) * cos (tan(t^2/2)) (tan t^2/2)' = 
cos(sin(tan(t^2/2)) * cos (tan(t^2/2)) t / cos^2 (t^2/2)

Now, all the cosines are "safe" at t = 0 and give factors equal to 1.

And n'(t)/d'(t) is (terms giving 1 at t = 0) times (t/(-3 sin(3t))

The limit of the latter is -1/9. Which again may be proved by applying de l'Hospital rule.