24yaar ankit.plz apna soln bhej.........aise confusion badhti jayegi....[2][2]
Q1\lim_{m\rightarrow infinity}[(\frac{e^{-100}.100^{100}}{100!})^{m}+5^{m}]^{1/m}
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24
1Q 2 .
A = lim (1.5n + [(1 + 1 /10000)10000 ]n)1/n
=lim ( 1.5n + [(1 + 1 + 10000*9999/2*10-8 + )]n)1/n
=lim ( 1.5n + [(1 + 1 + .49 + .16 + .04 + .008 )])n]1/n
= lim ( 1.5n + 2n)1/n
= lim 2( 1 + (1.5/2)n)1/n = 2 :)
1shit !! 1.5 is taken by me as 1 x 5 .. i thought u unnecessarily added 1 before 5 !!
11in the2nd 1 the 2nd term is becoming 2 (using binomial)
so
((1.5)n+2n)1/n (ur question needs to be edired)
(3/4n+2n)1/n
(3/4)(1+(8/3)n)1/n
so the answer is
2
24nahin yaar second ka to answer mere paas hai ....
ans 2 =2.......[1][1]
1Q 2 .
A = lim ( 5n + [(1 + 1 /10000)10000 ]n)1/n
=lim ( 5n + [(1 + 1 + 10000*9999/2*10-8 + )]n)1/n
=lim ( 5n + [(1 + 1 + .49 + .16 + .04 + .008 )])n]1/n
= lim ( 5n + 2n)1/n
= lim 5( 1 + (2/5)n)1/n = 5
// see post 21 & 19
1A = lim 5[1 + ((e-100. 100100)/(5 . 100!))m]1/m
ln A = ln 5 + 1/m ln[1 + ((e-100. 100100)/(5 . 100!))m]
ln A = ln 5 + 1/m ln[1 + ((e-100. 100100)/(5 . 100!))m] * ((e-100. 100100)/(5 . 100!))m / ((e-100. 100100)/(5 . 100!))m
ln ( 1 + x )/x = 1 .. . limit application
ln A = ln 5 + 1/m((e-100. 100100)/(5 . 100!))m
take lim ...1/m → 0 ...
so A = 5 .. Q1 .
24i dont know that..........if it is not given.....it shouldnt be.....
ok if u cant post the whoole soln then plz post ur main steps only.......or write in paint and upload ur picture....
1in question 1 → [...] simple bracket of greatest integer function ..
1i don't know how to use latex . .so typing the solution could take a lot of time ..
24Q2\lim_{n\rightarrow infinity}((1.5)^{n}+([(1+0.0001)^{10000}])^{n})^{1/n} where [.] denotes GINT.