1so in my solution i had actually found only the rhl.....what a foolish mistake.......
17 Answers
24@osama
I dont think I need to reply u on that....
bujt still for ur info ..I dont have Arihant book
This ques was from RSM ..
1@ Eureka 123
U COPIED BOTH QUESTION AND SOLUTION FROM ARIHANT'S BOOK
CHEATER , ..................HA HA HA HA ???
1thanks for reminding me....i will be more careful in the future
24tahts why I said ..find both LHL and RHL in subjective questions atleast..becoz in objective ques its not always necesaary..
124LHL=\lim_{h\rightarrow 0}[1+a^{-1/h}+(a+1)^{-1/h}+.....+(a+n-2)^{-1/h}]^{-h}
=> \lim_{h\rightarrow 0}\frac{1}{[1+a^{-1/h}+(a+1)^{-1/h}+.....+(a+n-2)^{-1/h}]^{h}}
=> LHL=1
RHL=\lim_{h\rightarrow 0}[1+a^{1/h}+(a+1)^{1/h}+.....+(a+n-2)^{1/h}]^{h}
=> (a+n-2)\lim_{h\rightarrow 0}[1+(\frac{1}{a+n-2})^{1/h}+(\frac {a}{a+n-2})^{1/h}+(\frac{a+1}{a+n-2})^{1/h}+.....]^{h}
=> RHL= a+n-2
hence LHL≠RHL..
=> Limt does not exist
1got an approach.first lets write :
lim (11+a1/x+(a+1)1/x+...+(a+n-2)1/x)x = z
x→0
now,the expression cn be written as : elog z and replace all x with 1/y . so x → 0 , y→ ∞
here, e lim.y→ ∞ ( log( 11 + ay + (a+1)y + .......... (a+n-2)y ) / y is of the form ∞ / ∞ .
now,cn put L-Hospitals rule to get d answer.
1yes but eureka told me that the answer was limit does not exists
1Is it not similar to http://www.targetiit.com/iit-jee-forum/posts/find-the-limit-10114.html ??
1but i have taken limit as 1/x tends to 0 and i agree that x tends to infinity that is the reason why in the 6th line terms less than 1 change to 0 as terms less than 1 to the power infinity are 0
1lim (1 + a^(1/x) + (a+1)^(1/x) +...+ (a+n-2)^(1/x))^x
x→0
replacing x by 1/x
we get
lim (1 + a^(x) + (a+1)^(x) +...+ (a+n-2)^(x))^(1/x)
1/x→0
taking (a+n-2)^x common from inside the backet we get
lim (a+n-2)(1/(a+n-2))^(x) + (a/(a+n-2))^x + ..... + 1^x)^(1/x)
now
1/a+n-2<1
a/a+n-2<1
and so on
so these terms to the power infinity ≈0
the only term left inside the brackett will be 1^x
so the final answer is (a+n-2)
hope that is the answer