#Solving a limit without L'hopitals Rule

60 messages · Page 1 of 1 (latest)

scenic horizon
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Hello, I know how to solve the limit using L'hopitals rule, but I'm trying to figure out how to solve the following limit using the given limits lim_x->0(sinx/x)=1 and lim_x->0(cosx-1/x)=0. The limit that needs to be solved is lim_x->0(sinx-sinxcosx/x^2). Could anyone help me with the steps?

terse irisBOT
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thorn groveBOT
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apex predator

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celest mulch
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@scenic horizon

scenic horizon
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doesn't x-x/x^2 still become an indeterminate form

celest mulch
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It’s 0/x^2

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So 0

scenic horizon
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wouldn't you plug 0 into x^2 and it would still be 0 as well?

summer harness
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have you tried

scenic horizon
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i tried factoring sinx out

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so it becomes sinx(1-cosx)/x^2

celest mulch
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0 goes to zero faster than x^2 does

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In the neighbourhood of 0

summer harness
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$\lim_{x\to 0}\frac{\sin x-\sin x \cos x}{x^2}=\lim_{x\to 0}\frac{1-\cos x}{x}$

thorn groveBOT
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Coffey, Slavic Taiga

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summer harness
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sorry for the error

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as sinx/x - > 1

scenic horizon
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no i haven't tried that yet

celest mulch
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Partial small angle approximation moment

scenic horizon
summer harness
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sin x ≈ x

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don't let anyone tell you otherwise

celest mulch
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Either taking out sinx/x or letting sinx=x

celest mulch
summer harness
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after that use the fact that 1-cosx = 1-(cos^2(x/2)-sin^2(x/2))=2sin^2(x/2)

summer harness
summer harness
celest mulch
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BA or BSc or BS or something

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sin(50BA)

summer harness
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Baa

scenic horizon
summer harness
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kinda

scenic horizon
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so i just sub x in for sinx?

summer harness
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see it for yourself

summer harness
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actually i advice against it

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substituting in stuff in a limit might lead to crucial errors

scenic horizon
celest mulch
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so lim f(x)h(x)g(x)=lim f(x) * lim g(x) * lim h(x) if the individual limits all exist

scenic horizon
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so (1) would be the lim f(x) and lim_x->0(1-cosx/x) would be lim h(x) right

scenic horizon
celest mulch
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because you said you're allowed to quote that (1-cosx)/x is zero

scenic horizon
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thats different from cosx-1/x=0 though right

celest mulch
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no

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-1*0=0

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one is just the negative of the other

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and negative 0 is 0

summer harness
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f->0 implies |f|->0 too

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not always the other way around

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but the two are equivalent if one of them -> 0 i think

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recheck this

dusty prairie
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@apex predator: sure of your proof. If f(x)=sin x - x, then the numerator writes f(x)-0.5 f(2x) which tends to 0, but you don’t know its rate of convergence compared to x² (except if you know Taylor series which seem to be out of the scope).

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@tommy if you use the definition of the derivative for sin and cos you don’t use the Hospital rule per se

dusty prairie
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sin x/x = (sin x - sin 0)/(x -0) so lim in 0 is cos 0=1
Don’t write cos x - 1/x but (cos x - 1)/x, which equals (cos x - cos 0)/(x - 0) so lim in 0 is -sin 0=0.

celest mulch
celest mulch