Solve the question given in attachment..................................................

let the limit be L. take ln on both sides 
so lnL= Lt (n+1)^2*ln(cosln[(n-1)/(n+1)])
now put n+1=1/x so that x tends to zero.
lnL= Lt ln(cosln[1 – 2x])/x^2
now ln(cosln[1 – 2x])/x^2= {ln(cosln[1 – 2x])/(cosln[1 – 2x] – 1)}*{[cosln[1 – 2x] – 1]/ln^2((1-2x)/2)}*[ln((1 – 2x)/2)/x]^2
now although this appears to be really messed up, but turns out that the individual limits ln(cosln[1 – 2x])/(cosln[1 – 2x] – 1), [cosln[1 – 2x] – 1]/ln^2((1-2x)/2) and [ln((1 – 2x)/2)/x]^2 all are easily solvable by putting y= cosln[1 – 2x] – 1, z= ln((1-2x)/2 and using cost – 1= – 2sin^2(t/2), and u= – 2x respectively.