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Integral Calculus

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28 Feb 2007 16:42:45 IST

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740

a prob..........pls solve this quickly.

None

_{[x ]}_{[ ]} (2^x+ 3^x) / (3^x+ 1)

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vasanth_mech pilani

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28 Feb 2007 16:44:31 IST

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can u b more clear

like use ( ) [ ] { }

CyBorG

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28 Feb 2007 16:48:28 IST

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I hope this is your question lim(x~inf)(2^x+3^x)/(3^x+1) It is inf/inf form

=lim(x~inf)(2^xlog2+3^xlog3)/(3^xlog3) [L Hospital's rule]

=lim(x~inf)((2/3)^xlog2/log3+1)

=1 [Since 2/3<1 so (2/3)^infinity tends to zero]

Hope you are clear!

sneha kulkarni

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28 Feb 2007 16:48:42 IST

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ok i'll try to say it in words..............

find the limit if

limit as x tends to infinity (2^x + 3^x) divided by (3^x + 1)

pls do tell me if it's not clear again........

sneha kulkarni

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28 Feb 2007 16:51:56 IST

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heyy KAB i dint understand it. i dont know hospital's rule.........i am only in 11th now..........that too in state syllabus...please make it clear......by using the simple standard results.........

vasanth_mech pilani

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28 Feb 2007 16:57:05 IST

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there are 7 indeterminate forms like 0/0 , infinity/infinity,........

when u get these terms insidethe lim by direct substitution

u differentiate the quantity inside the limit and reduce it to a determinate form, if u still dont get it u continue to differentiate.......

this is l'hospital's rule........

vasanth_mech pilani

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28 Feb 2007 16:57:46 IST

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hope u're clear

dont mind to ask if u're not clear

just dont forget to rate me.......

CyBorG

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28 Feb 2007 16:58:55 IST

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Sorry!

Here is another method.

Take3^x common from numerator and denominator.

So,

lim(x~inf) ((2/3)^x+1)/(1+(1/3^x))

Now see that 2/3<1 So (2/3)^infinity tends to zero .Also (1/3)^infinity tends to zero

So required answer is (0+1)/(1+0)=1

Now I think you are clear.

vasanth_mech pilani

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28 Feb 2007 17:00:08 IST

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i agree with u KAB

u get ma salute

sneha kulkarni

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28 Feb 2007 17:05:49 IST

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hi vasanth..............pls tell me cant we use any of the following forms...........i tried but i am not getting the further simplification..............

here are the standard results.........

lim(n-->infinity) ( 1 + (1/n))ⁿ= e

lim(n-->infinity) (1 + (x/n))ⁿ= e^x

^{pls tell me ever if i am wrong........}

saarika .

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28 Feb 2007 17:08:44 IST

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try this method .take 3^x common from both numerator& denominator

infinity [ (2^x/3^x)+1]/1+(1/3^x)as 2/3&1/3 are proper fractions,(2/3)^xtends to 0 as x tends to infinity. hence ans is (0+1)/(1+0)=1

vasanth_mech pilani

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28 Feb 2007 17:13:06 IST

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i dont think its possible for this sum

even if possible it'd be the longer way around

always try to complete it as simple as u can .......

sneha kulkarni

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28 Feb 2007 17:13:47 IST

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i am much clear now............... thank you adarsh, vasanth and saarika.........

BHARGAV CHINTAKUNTA

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5 Mar 2007 18:25:46 IST

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i know this answer but i missed salute

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