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Differential Calculus

Blazing goIITian

 Joined: 8 Dec 2006 Post: 351
20 Jan 2007 23:25:05 IST
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20
1980
let's see who solves this..
Engineering Entrance , JEE Main , JEE Advanced , Mathematics , Differential Calculus

lim             (cosa)^x-(sina)^x+sin2a/x-4
x->4
the  whole expression is divided to x-4..

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Hot goIITian

Joined: 18 Dec 2006
Posts: 146
21 Jan 2007 10:04:04 IST
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I think numerator is finite at x=4 and denominatr is zero at x=4

Hot goIITian

Joined: 5 Dec 2006
Posts: 169
21 Jan 2007 12:17:28 IST
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NISHANT , I THINK APPLYING L HOSPITALS RULE SOLVES IT .

Blazing goIITian

Joined: 26 Oct 2006
Posts: 530
21 Jan 2007 12:24:56 IST
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by applying l hospital rule i m getting............................
(cosa)^4 logcosa - (sina)^4 logsina + sina/2
is the answer correct.............plz tell me

Forum Expert
Joined: 6 Jan 2007
Posts: 706
21 Jan 2007 12:25:26 IST
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But it is not of 0/0 form.So how can we apply L'Hospitals rule.

Hot goIITian

Joined: 20 Jan 2007
Posts: 122
21 Jan 2007 15:58:19 IST
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hi nishant, the question given by u is incomplete.
if u simply put x=4 u get sin2a+cos2a in numerator & 0 in the denominator
if the numerator is positive quantity and limit is approaching from RHS then ans is +infinity and if the numerator is negative quantity and lmit is approaching from LHS then ans is -infinity as the interval of a is not defined by u so it does not give any proper solution.

Blazing goIITian

Joined: 8 Dec 2006
Posts: 351
21 Jan 2007 17:44:27 IST
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well,sorry for not telling the interval to which a belongs.it is(0,/2)

Hot goIITian

Joined: 5 Dec 2006
Posts: 169
21 Jan 2007 20:29:43 IST
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nrki99, you failed in checking continuity of the limit before explaining the answere, i think.

Hot goIITian

Joined: 20 Jan 2007
Posts: 122
22 Jan 2007 03:59:40 IST
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hi gkp, i think u can solve this problem better than me so
u can solve this problem.

message to nishant : is u made this problem ?
if yes, don't try make problems because there is already many problems
if  no, then tell the source from which u got this problem.

Blazing goIITian

Joined: 8 Dec 2006
Posts: 351
22 Jan 2007 17:31:08 IST
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i haven't made this problem.i am no genius to make problems.its frm a magazine called mathematics today..plz try to get the answer..

Forum Expert
Joined: 6 Jan 2007
Posts: 706
22 Jan 2007 17:56:32 IST
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Try to take (cosa)x-4-(sina)x-4  outside so that we get
(((cosa)x-4-1)-((sina)x-4-1))/(x-4) and then simplify. I am not sure whether it will work or not.

New kid on the Block

Joined: 19 Jan 2007
Posts: 13
23 Jan 2007 00:48:42 IST
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numerator will be cos2a+sin2a.
denominator will be 0.
so no l'hospital's rule.
the ans is infinite.
infinite is also a limit,
for eg., limit tanx as x->90- is infinte.
not a gud problem

Hot goIITian

Joined: 20 Jan 2007
Posts: 122
23 Jan 2007 03:52:31 IST
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hi rajiv, don't be oversmart, u didn't know whether the numerator is positive or
negative so before solving problem lets first read what argument is given by others

Blazing goIITian

Joined: 8 Dec 2006
Posts: 351
23 Jan 2007 17:28:40 IST
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rajiv ,my friend.its cosa^x and sina^x.u can't write that as cos2a..

Blazing goIITian

Joined: 8 Dec 2006
Posts: 351
23 Jan 2007 23:57:01 IST
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KAB,i tried ur method but couldn't solve the question that way.
hope someone else comes up with a soln. quickly.
i wish some expert could help..

Scorching goIITian

Joined: 31 Dec 2006
Posts: 291
24 Jan 2007 01:00:34 IST
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Let f (x) =[ (cosa)^x - (sina)^x + sin2a ] / (x-4)

=[ (cosa)^x - (sina)^x - cos2a ] / (x-4)  + [sin2a + cos2a]/(x-4)

= g(x)  +  h(x)  { let }
Then,
lim ( x--> 4 )    g(x) is of 0/0 form.     ok?

use l'hospital's rule and find the value of g(x) .It is a finite value.

Now, f (x) = something finite + h(x)

And I think it is clear that h(x) is of form    something finite/infinity

Therefore , lim ( x--> 4 )  f (x) doesn't exist

Though I believe there was no need to do this all, putting 4 in the expression of f(x) explains it as told by Malay.

Blazing goIITian

Joined: 8 Dec 2006
Posts: 351
25 Jan 2007 10:33:39 IST
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thank u all for providing answer to my question..

Hot goIITian

Joined: 20 Jan 2007
Posts: 122
26 Jan 2007 05:27:42 IST
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hey aakriti
the solution given by u is incomplete
because u consider the some finite value/ 0 =infinity
because some +ve finite value/0 = +infinity & some -ve finite value/ 0= -infinty
here may be the cases of 0+ and 0-. i am also very confused regarding that
problem.so u must use the interval here for solving the problem
i think the problem is wrong!
so nishant i advice u to ask this problem from any expert

Cool goIITian

Joined: 31 Dec 2006
Posts: 31
26 Jan 2007 08:27:46 IST
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cant we directly substitute x=4 in the expression  and since cos and sin has value btwn 0,1 so (cosa)^x-(sina)^x+sin2a/x cant be equal to 4 therefore cant be nethier indeterminate i think ans is infinite

Blazing goIITian

Joined: 8 Dec 2006
Posts: 351
26 Jan 2007 13:06:19 IST
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will some forum expert kindly reply to my question........

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