trig question (1 Viewer)

[Constip8edSkunk]

are these methods correct?

lim_x-->0 (cosec x - cot x)


1/
= lim_x-->0 [(x/x) (1/sinx - 1/tanx)]
= lim_x-->0 [(1/x) (x / tanx - x / sinx)]
= lim_x-->0 (x / tanx - x / sinx) * lim_x-->0 (1/x)
= (1-1) * lim_x-->0 (1/x)
=0



2/
= lim_x-->0 [( sinx / x)( tanx / x)(1/sinx - 1/tanx)]
= lim_x-->0 [ tanx / x² - sinx / x²]
= lim_x-->0 [(1/x) (sinx / x - tanx / x)]
= (1-1) * lim_x-->0 1/x
= 0

thx

 

[McLake]

No.

The lim as 1/x -> 0 is "infinity"
0*"infinity" is undefined ...

lim (x->0) (cosec x - cot x)
= lim (x -> 0) (1/sinx - cosx/sinx)
= lim (x -> 0) ((1 - cosx)/sinx)
= 0/0

WAY 1:
(UNI STUFF)
let f(x) = 1 - cosx
let g(x) = sinx
so lim (x -> 0) f(x)/g(x)
= lim (x -> 0) f'(x)/g'(x)
= lim (x -> 0) sinx/cosx
= 0/1
= 0
using Le Hopitals Rule
lim (x -> 0) f(x)/g(x) = lim (x -> 0) f'(x)/g'(x)
so lim (x->0) (cosec x - cot x) = 0

WAY 2:
I don't know ...

 

[wogboy]

Yep there's an interesting rule called L'Hopitals Theorem (you'll learn at uni if you do maths) that goes like this:

If you have 2 continuous functions f(x) and g(x), and f(a)=g(a)=0 (or also if f(a) and g(a) are both infinite, having vertical asymptotes at x=a), then:

limit (x->a) f(x)/g(x) = lim (x->a) f'(x)/g'(x),

====================================

Another way to do that question (without using L'Hopital's rule) is:

lim (x->0) cosecx - cotx
= lim(x->0) (cosecx - cotx) * (cosecx + cotx) / (cosecx + cotx)
= lim (x->0) (cosec^2(x) - cot^2(x))/(cosecx + cotx)

but cosec^2(x) = 1 + cot^2(x) (trig identity)
so cosec^2(x) - cot^2(x) = 1

so the limit becomes:
= lim(x->0) 1/(cosecx + cotx)
= lim(x->0) sinx/(sinx*cosecx + sinx*cotx)
= lim(x->0) sinx/(1 + cosx)
= 0/(1 + 0)
= 0

The above two methods of yours are wrong for solving these limits, since when you evaluate you limit (by actually subbing in 0 as an x-value) you need to be careful you don't get infinity multiplied by zero (or zero divided by zero for that matter)

 

[Constip8edSkunk]

ok cool thanx