IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

snehavenus

_{[x ]}

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

vasanth

can u b more clear

like use ( ) [ ] { }

KAB

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!

snehavenus

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........

snehavenus

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

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

hope u're clear

dont mind to ask if u're not clear

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

KAB

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

i agree with u KAB

u get ma salute

snehavenus

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

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

x

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

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 .......

snehavenus

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

cbg2007

i know this answer but i missed salute

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