11Kumar ...............................................
question recheck kar lo JI
15 Answers
11kumar kindly write the question again
its not clear
square upper hai ya niiche??????????
and wat r a, b,c
11
1\lim_{x\rightarrow o}\left( \frac{a^{x}+b^{x}+c^{x}}{3}}\right)^{\frac{2}{x}}
=\huge e^{\lim_{x\rightarrow 0}\left(\frac{2}{3} \right)\frac{a^{x}+b^{x}+c^{x}-3}{x}}
=\huge e^{\left(lnabc \right)^{\frac{2}{3}}}
106it is = 2cos(pi/4n)sin(pi/4n)*n/2
= sin(pi/2n)*pi/4(pi/2n)
= taking limits... n-->∞ ==> pi/4n --> 0 ..
= pi/4
... calculation mistakes possible
1lim n→∞ ncos(p∩/4n)sin(∩/4n)
let n=1/y
therefore
lim y→0 1/y X cos(y∩/4) X sin(y∩/4)
multiplying and dividing by y∩/4 we have
lim y→0 y∩/4y X cos(y∩/4)
=∩/4
answer=∩/4
62This is the first step
\huge e^{(\lim_{x\rightarrow 0}\left(\frac{2}{3} \right)\frac{a^{x}+b^{x}+c^{x}-3}{x})}
now this canbe writtenas
\huge e^{(\lim_{x\rightarrow 0}\left(\frac{2}{3} \right)\frac{a^{x}-1+b^{x}-1+c^{x}-1}{x})}
look at \lim_{x\rightarrow 0} ((a^{x}-1)/x+(b^{x}-1)/x+(c^{x}-1)/x))
which is \ln a + \ln b + \ln c
now this is ln(abc)
rest you can work out
1what is the asnwer for second sum i.e post no 7 and i solved it in post no 10
11soory amit but ur solution is wrong[2]
lim n→∞ ncos(p∩/4n)sin(∩/4n)
let n=1/y
therefore
lim y→0 1/y X cos(y∩/4) X sin(y∩/4)
multiplying and dividing by y∩/4 we have
lim y→0 y∩/4y X cos(y∩/4)
=∩/4
answer=∩/4
TELL ME 1 THING HOW WOULD U MULTIPLY AND DIVIDE BY y WHEN IT TENDS TO 0
JUSTIFY THE BOLDED STEP!!!!!!!!!!!
1why cant we multiply numerator and denominator by the same number?
11U CAN NEVER MULTIPLY O OR ∞ FROM ANY EQUATION
IT DOES NOT MAKES ANY SENSE
AND IT IS AGAINST THE RULES OF THE MATHEMATICS ( AS FAR AS I KNOW )[1]