Question

please any body can solve this.. lim x->0 cosecX-cotX/x

Answer 1

\[\lim_{x \rightarrow 0}\frac{\csc x - \cot x}{x}\]
Familiar with L'Hopital's Rule?

Answer 2

whats is l hospital rule?

Answer 3

on second thought, I may have been hasty... forget about that :)
Just focus on
csc x - cot x

Notice anything?

Answer 4

cant apply this..

Answer 5

Yeah, I realised (a bit too late :) )
But focus on
csc x - cot x

Answer 6

whats i got?

Answer 7

\[\csc x - \cot x = \frac{1}{\sin x} - \frac{\cos x}{\sin x}=\frac{1-\cos x}{\sin x}\]
Look familiar?

Answer 8

please tell me full steps how to solve this. i dont know L hostipal rule and i cant apply that

Answer 9

Okay, we might not be able to use L'Hopital, but maybe we can be sneaky with it :)
First things first, though, to make it simple, carrying on from that
csc x - cot x bit:
\[\frac{1-\cos x}{\sin x}=\sqrt{\frac{(1-\cos x)^{2}}{\sin^{2}x}}=\sqrt{\frac{(1-\cos x)(1-\cos x)}{1-\cos^{2}x}}\]
\[=\sqrt{\frac{(1-\cos x)(1-\cos x)}{(1-\cos x)(1+\cos x)}}\]
cancelling out 1-cos x
we get\[=\sqrt{\frac{1-\cos x}{1+\cos x}}\]
Does it look familiar now?

Answer 10

ok than?

Answer 11

\[\sqrt{\frac{1-\cos x}{1+\cos x}}=\tan \left( \frac{x}{2} \right)\]
Can you do it from here?

Answer 12

tell me please how can i solve that question..

Answer 13

What have we done so far?
We've shown that
\[\csc x - \cot x = \tan \left( \frac{x}{2} \right)\]
right?

Answer 14

ok

Answer 15

For simplicity, we can let
\[f(x)=\tan \frac{x}{2}\]
so we can write \[\lim_{x \rightarrow 0}\frac{\csc x - \cot x}{x}=\lim_{x \rightarrow 0}\frac{f(x)}{x}\]
Catch me so far?

Answer 16

look i got \[\lim_{x \rightarrow o}{\frac{ 1 }{ \sin }-\frac{ cosx }{ sinx }}\]

Answer 17

ok than?

Answer 18

And you'd end up with an indeterminate form... which would require the use of L'Hopital's rule... step by step, and here we go:
Refer to my previous message:
\[f(x)=\tan \frac{x}{2}\]
So your limit is
\[\lim_{x \rightarrow 0}\frac{\csc x - \cot x}{x}=\lim_{x \rightarrow 0}\frac{f(x)}{x}\]
Getting it so far? The 'tricky' part comes next...

Answer 19

than?

Answer 20

What is f(0)?

Answer 21

you mean answer is zero 0?

Answer 22

No, but f(0) is 0, right?

Answer 23

No. terenzreignz is asking what\[\tan(\frac{0}{2})\]is

Answer 24

i think its zero?? am i right?

Answer 25

This is my most hated question and subject

Answer 26

Indeed
\[f(0)=\tan\frac{0}{2}=\tan 0 = 0\]
So...
f(x) = f(x) - f(0)
RIGHT?
I mean, subtracting zero from a function won't affect it, right?

Answer 27

yeh

Answer 28

Okay, on another note
x = x - 0
This is obvious right?

Answer 29

Prove: sinx=1+cosx

Answer 30

yeh its rite but answer is not zero//

Answer 31

sin x is not equal to 1 + cos x.
Anyway
We have established that
f(x) = f(x) - f(0)
and
x = x - 0
THEREFORE
\[\lim_{x \rightarrow 0}\frac{f(x)}{x}=\lim_{x \rightarrow 0}\frac{f(x)-f(0)}{x-0}\]
This must look familiar to you by now...

Answer 32

@terenzreignz it would be nice if u made one clear post. i dont get this problem and i very much would like to

Answer 33

ok

Answer 34

ok than whats the answer?

Answer 35

tell me the answer please i cant understand it?

Answer 36

Have you taken up derivatives?

Answer 37

you can just post the worked out solutions you know. later we can ask questions if we dont get any parts...

Answer 38

my ideas is :
lim (x->0) (csc(x) - cot(x))/x
= lim (x->0) [1/sin(x) - cos(x)/sin(x)]/x
= lim (x->0) (1- cos(x))/(x*sin(x))
now, u can use the identity :
1-cos(x) = 2 sin^2 (x/2), so it can be
= lim (x->0) 2 sin^2 (x/2)/(x*sin(x))
= lim (x->0) 2 sin(x/2) sin(x/2) /(x*sin(x))
and finally, u can use these formulas to find the value that limit
lim (x->0) sinAx/Bx = A/B and also
lim (x->0) sinAx/sinBx = A/B
good luck

Answer 39

the answer in my book is 1/2 but how?

Answer 40

yes, correct. the answer is 1/2

Answer 41

can you show me the steps please??

Answer 42

i have showed u above, u just understanding my steps and i hope u can continue it and to get the answer

Answer 43

This is a summary. We started with
\[\lim_{x \rightarrow 0}\frac{\csc x - \cot x}{x}\]
Now it can be shown that
\[\csc x - \cot x = \tan \frac{x}{2}\]
So it becomes
\[\lim_{x \rightarrow 0}\frac{\tan\frac{1}{2}x}{x}\]
But
\[\tan \frac{1}{2}(0) = 0\]
meaning it can be subtracted from the numerator without changing its value.
Also, 0 can be subtracted from the denominator, without changing its value. The limit then becomes\[\lim_{x \rightarrow 0}\frac{\tan\frac{1}{2}x -\tan\frac{1}{2}(0)}{x - 0}\]
But this is just the derivative of tan(x/2) evaluated at x = 0

\[\huge =\frac{d}{dx}\left[ \tan\frac{1}{2}x \right]_{x=0}\]
\[\huge \frac{d}{dx}\tan\frac{1}{2}x=\frac{1}{2}\sec^{2}\frac{1}{2}x\]
Now just evaluate it at x = 0

Answer 44

i cant///

Answer 45

look the last equation is
lim (x->0) 2 sin(x/2) sin(x/2) /(x*sin(x))
= lim (x->0) 2 sin(x/2)/x * sin(x/2)/sinx
= 2 * (1/2)/1 * (1/2)/1
= 1/2

Answer 46

how i cant solve it..

Answer 47

radEn can you do it more easy to understand>?

Answer 48

hellow any body here to solve it?

Answer 49

Read the summary.

Answer 50

i cant

Answer 51

i cant understand any thing//

Answer 52

how tan=sec

Answer 53

\[\tan(x) \neq \sec(x)\]However,\[\frac{d}{dx}\tan(x)=\sec^{2}(x)\]When combined with the chain rule:\[\frac{d}{dx}f(g(x))=f'(g(x)) \times g'(x)\]We can say that\[\frac{d}{dx}\tan(\frac{x}{2})=\frac{1}{2}\sec^{2}(\frac{x}{2})\]

Answer 54

hahaha i cant understand it again..