Fundamental trig ratio limit (1 Viewer)

[nick1048]

please help with the following two questions if you can:

1) lim (sin^2 x) / x
x-->0



2) lim (1 - cos x) / x^2
x-->0


The textbook is very vague in providing the reasons as to why these instances occur, hence my lack of understanding the algebra lol. Thanks in advance to whoever can get these damn things solved.

 

[Estel]

bumping after 15 min?
pfft

1) lim (sin^2 x) / x
x-->0
= lim x --->0 sinx/x . sinx
= 1. lim x --->0 sinx
= 0

2) lim (1 - cos x) / x^2
x-->0
= lim x-> 0 (2sin^2[x/2])/x^2
= lim x->0 2.[sin(x/2)/x]^2
= 1/2

 

[nick1048]

I dont understand how you work the algebra...

 

[Slidey]

Give me L'Hospital's rule any day.

 

[nick1048]

bumping after 15 min?
pfft

1) lim (sin^2 x) / x
x-->0
= lim x --->0 sinx/x . sinx
= 1. lim x --->0 sinx
= 0

2) lim (1 - cos x) / x^2
x-->0
= lim x-> 0 (2sin^2[x/2])/x^2
= lim x->0 2.[sin(x/2)/x]^2
= 1/2

ok i get number 1 because sinx as x approaches 0 would be sin0 = 0... but the second one throws me off... I don't understand how you get from 1 - cosx to 2sin(x/2)

 

[FinalFantasy]

he just used cosx=1-2sin²(x\2)

 

[Estel]

Give me L'Hospital's rule any day.

Not even safe to use in 4u =p

 

[Trev]

Give me L'Hospital's rule any day.

And that would be? A French hospital?

 

[FinalFantasy]

i heard from slide rule that that "hospital" rule is
when u got lim of x--->0 .............
and it's over x, i.e u can't divide by 0
then u can find lim x-->0 the derivative of the numerator over the derivative of the denominator

 

[Slidey]

It would be the name of the French mathematician which the rule is named after.

Basically, for limits which BOTH numerator and denominator give zero or infinity, you can find the true limit by this method: For
lim f(x)/g(x)
x->a
Then instead find
lim f'(x)/g'(x)
x->a

For example:

lim sinx/x
x->0
=lim cosx/1
x->0
=1

And:
2) lim (1 - cos x) / x^2
x-->0
=lim sinx/2x
x->0
=1/2

BE CAREFUL. For example, why would the following limit produce an incorrect answer if you applied l'Hospital's rule?

lim (2 - cos x) / x^2
x-->0

It is because the top 2-cosx is not equal to zero or infinity when x=0.

In this case you cannot use l'Hospital's rule.

Further, for limits such that
lim f'(x)/g'(x)
x->a
Again produces an indeterminate answer and meets the conditions for l'Hospitals rule, a second application of l'Hospital's rule is used. That is, find:
lim f''(x)/g''(x)
x->a

 

[KFunk]

It would be the name of the French mathematician which the rule is named after.

Basically, for limits which BOTH numerator and denominator give zero or infinity, you can find the true limit by this method: For
lim f(x)/g(x)
x->a
Then instead find
lim f'(x)/g'(x)
x->a

Nifty......

 

[acmilan]

Please refrain from bumping nick1048 if you're not getting a response, it would just be that no one is online that can help, not that they cant see the question.

 

[Estel]

But you already got an answer, and an explanation of the part you didn't answer by someone else.
Bumping every 15 min is unfair and a waste of time. If there is something you still don't get, then please ask in respect to that.

 

[Slidey]

Not even safe to use in 4u =p

Says who?