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Differential Calculus

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16 Sep 2007 21:21:08 IST

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858

Confusing Question on LIMITS

None

Evaluate

1) Lim cos (1/x)

x-->0 -----------------------

cos (1/x)

2) Lim 2+ cos(1/x)

x-->0 ----------------------

2+ cos(1/x)

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Comments (7)

siva viper

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Joined: 15 Apr 2007

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16 Sep 2007 23:42:59 IST

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isn't the answer equal to 1??????

venkat v.t

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17 Sep 2007 00:15:03 IST

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Yes.In both the problems the limit is 1.
Before applying limit u should simplify the function.
Here x tends to 0 means x reaching to zero but not equal to zero.
So we can cancel numerator and denominator as they are equal.
Therefore the limit is 1.

Avinash Sharma

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Posts: 259

17 Sep 2007 02:25:04 IST

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Hello rahulkarmakar14

venkat_tatolu has given the clear n correct answer. You should keep the following rules in mind while solving the questions on limits :

1. _{[ x][ p]} [f(x) + g(x)] = _{[ x][ p]} f(x) + _{[ x][ p]}g(x)

2. _{[ x][ p]} [f(x) - g(x)] = _{[ x][ p]} f(x) - _{[ x][ p]}g(x)

3. _{[ x][ p]} [f(x) . g(x)] = _{[ x][ p]} f(x) . _{[ x][ p]}g(x)

4. _{[ x][ p]} [f(x) / g(x)] = _{[ x][ p]} f(x) / _{[ x][ p]}g(x)

wander wath

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17 Sep 2007 17:11:30 IST

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1. we get lim ( 1)=1
x--0

2. lim (1)=1
x--0
rate if convinced

Gaurav |spideyunlimited| Ragtah

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17 Sep 2007 17:21:32 IST

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simple answer !
1 for both

nityesh sharma

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19 Sep 2007 15:05:42 IST

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the answer is clearly 1. After simplifying the question the numerator and denominator simply cancel with each other and the answer is 1.

Shikhar Shukla

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22 Sep 2007 09:34:10 IST

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in both the cases,_{[ x}_]

_{[ 0]} cos1/x can vary b/w -1 to 1 but the vaue of num n den being the same will get cancelled to give 1.

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