How do I do this? Calculus limit inf-inf form with L'Hopital's Rule! Reddit r/calculus

Learn how to evaluate the limit of ln(1-cos(x))-ln(x^2) as x goes to infinity with L'Hospital's rule. We first need to combine the logarithms together and then work out the limit inside out. Subscribe for more calculus tutorials.

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[bprpcalculusbasics]

Calculus 1, how do you evaluate these limits (not 0/0 form) Reddit r/homeworkhelp

[alex_ramjiawan]

I'm aware of the inside-out changeability due to ln being a continuous function, but I never learned why that's the case.

[blueslime]

1-cosx ~ x²/2 (Taylor series) therefore we have
ln(x²/2)-ln(x²)=ln½, done

[Brid]

for some reason i did the following:
lim {ln(1-cosx) - ln(x^2)} as x->0
lim [ln{(1-cosx)/x^2}] as x->0
let L = lim [ln{(1-cosx)/x^2}] as x->0
e^L = lim (1-cosx)/x^2 as x->0
now to focus on the RHS
lim {(1-cosx)/x^2} as x->0
we get 0/0, so we use L'Hopital's rule
after which, we get,
lim (sinx/2x) as x->0
=(1/2)*{lim (sinx/x) as x->0}
the limit in the latter portion equals 1
=1/2
so e^L=1/2
L=ln(1/2) = -ln(2)

[AltAaltonnov]

Why can't you apply L'Hospital to lim x-> 0 sin (x) / x? It's of the form 0/0, and L'Hospital gives cos x / 1 and cos(0)/1 equals 1....

[jackychanmaths]

Since lim_(x->0) sin x / x cannot be solved by L' Hospital's rule, lim_(x->0) [(1 - cos x)/x^2] also cannot be solved by L' Hospital's rule since the derivative of cos x also depends on lim_(x->0) sin x / x.

The correct way should be
lim_(x->0) [(1 - cos x)/x^2]
= lim_(x->0) {[2sin^2 (x/2)]/x^2}
= 1/2* lim_(x->0) [sin (x/2) / (x/2)]^2
= 1/2* { lim_(x->0) [sin (x/2) / (x/2)] }^2
= 1/2 * 1^2
= 1/2

[Mediterranean]

The limit of 1-cos (x)/x^2 is a famous limit (=1/2)