What is #lim_(nrarroo)(-1)^(n-1)sin(pisqrt(n^2+0.5n+1))# ?

3 Answers
Mar 25, 2016

The limit is #-1/sqrt2#

Explanation:

#" " #

We need to find the asymptote to #sqrt(n^2+0.5n+1)#.
(Thanks, George C.)

To find (linear) oblique asymptote #y=mn+b#,

#m = lim_(nrarroo) sqrt(n^2+0.5n+1)/n = 1#.

Now we find #b#:

#b= lim_(nrarroo) sqrt(n^2+0.5n+1)/mn = lim_(nrarroo) sqrt(n^2+0.5n+1)/n = 1/4#.

So, #sqrt(n^2+0.5n+1)# is asymptotic to #n+1/4#.

Alternative method from George C

#sqrt(n^2+0.5n+1) = sqrt((n+1/4)^2+15/16)#

# = (n+1/4)sqrt(1+15/(16(n+1/4)^2)#

(If I've deduced the method correctly, our goal is to get #(mn+b)sqrt(1+g(n))# where #lim_(nrarroo)g(n)=0#.)

Returning to the big question

#lim_(nrarroo)(-1)^(n-1)sin(pisqrt(n^2+0.5n+1)) = lim_(nrarroo)(-1)^(n-1)sin(pi(n+1/4)sqrt(1+15/(16(n+1/4)^2)))#

For even #n#, say #n=2k#,
we get #-sin(2kpi+pi/4) = -sin(pi/4) = -1/sqrt2#

For odd #n#, say #n=2k+1#,
we get #sin(2kpi+pi+pi/4) = sin((5pi)/4) = -1/sqrt2#

Therefore,

#lim_(nrarroo)(-1)^(n-1)sin(pisqrt(n^2+0.5n+1)) = -1/sqrt2#

Mar 25, 2016

#-sqrt2/2#

Explanation:

Extra care should be taken since we are dealing with a periodic function of the kind
#f(x)=+-sinx#

It's not sufficient that, when #n->oo#, #n^2# is much greater than #0.5n+1#, because if #sqrt(n^2+0.5n+1)-n > 0# than the function will have a value possibly different than zero.

Intuitively it's easy to see that #sqrt(k+1)->sqrt k# (or, even more importantly for the case, that #sqrt(k+1)-sqrt k->0#) when #k->oo#
Or
#sqrt(1001)-sqrt(1000)=0.015,807#
and
#sqrt(1,000,001)-sqrt(1,000,000)=0.000,500#
and so on.

But let's prove that #sqrt(n^2+0.5n+a)->sqrt(n^2+0.5n)# when #n->oo#, where #n in NN# and #a in RR#
#lim_(n->oo) (sqrt(n^2+0.5n+a)-sqrt(n^2+0.5n))*((sqrt(n^2+0.5n+a)+sqrt(n^2+0.5n)))/(sqrt(n^2+0.5n+a)+sqrt(n^2+0.5n))=#
#=lim_(n->oo) (cancel(n^2)+cancel(0.5n)+a-cancel(n^2)-cancel(0.5n))/(sqrt(n^2+0.5n+a)+sqrt(n^2+0.5n))#
#=lim_(n->oo) a/(sqrt(n^2+0.5n+a)+sqrt(n^2+0.5n))#
#=lim_(n->oo) a/oo=0#

Since the value of #a# (in the present case #a=1#) doesn't matter to the result, we can make
#lim_(n->oo) sqrt(n^2+0.5n+1)=lim_(n->oo) sqrt(n^2+0.5n+1/16)=lim_(n->oo) sqrt((n+1/4)^2)=lim_(n->oo) (n+1/4)#

So the main expression becomes

#=lim_(n->oo) (-1)^(n-1)sin[pi(n+1/4)]#
#=lim_(n->oo) (-1)^(n-1)sin(n*pi+pi/4)#
But
#sin(n*pi+pi/4)=sin(n*pi)*cos(pi/4)+sin(pi/4)*cos(n.pi)=#
(remembering that #n in NN# => #sin(n*pi)=0#)
#=sqrt2/2*cos(n*pi)#
In the main expression
#=lim_(n->oo) (-1)^(n-1)*sqrt2/2*cos(n*pi)#

Now consider the possible values of the expression above
If n is odd #=> (-1)^("even number").sqrt2/2*cos[("odd number")*pi]=1*sqrt2/2(-1)=-sqrt2/2#
If n is even #=> (-1)^("odd number").sqrt2/2*cos[("even number")*pi]=(-1)*sqrt2/2*1=-sqrt2/2#

So, for #n in NN#
#=lim_(n->oo) (-1)^(n-1)*sqrt2/2*cos(n*pi)=-sqrt2/2#

Mar 27, 2016

#-1/sqrt2#.

Explanation:

I have revised my answer, thanks to George for his affirmation on his answer. I am sorry for having overlooked some nicety in applying the notion of 'one is asymptotic with another', in this limit problem.

The correction follows, without changing my approach.

#pisqrt(n^2+0.5n+1)=npisqrt(1+0.5/n+1/n^2))=npi(1+(1/2)(0.5/n)+O(1/n^2))#, using expansion in powers of #1/n.#
Here, #ntooo# through integer values 1,2,3,..,

The limit is same as the limit for
#(-1)^nsin(npi+pi/4+O(1/n)) =-1/sqrt2#, as given in the other answers..