12 - Limit Continuity and Differentiability Questions Answers

I think n is printed by mistake, so considering it as x

the question will be limit x tends to ∞

x{ [tan^-1 (x+1/x+4)] - (π/4)}

consider [tan^-1 (x+1/x+4)] - (π/4) = θ

so [tan^-1 (x+1/x+4)] = (π/4)+θ

so (x+1/x+4) = tan(π/4+θ)

or x = (-3-5tanθ)/2tanθ

so converted question will be limθ->0  (-3-5tanθ)θ/2tanθ

solve

I think n is printed by mistake, so considering it as x

the question will be limit x tends to ∞

x{ [tan^-1 (x+1/x+4)] - (π/4)}

consider [tan^-1 (x+1/x+4)] - (π/4) = θ

so [tan^-1 (x+1/x+4)] = (π/4)+θ

so (x+1/x+4) = tan(π/4+θ)

or x = (-3-5tanθ)/2tanθ

so converted question will be limθ->0  (-3-5tanθ)θ/2tanθ

now we know that limθ->0 tanθ/θ = limθ->0 θ/tanθ = 1

so next line will be limθ->0 (-3-5tanθ)/2 = -3/2 

We have to calculate limit n -> ∞ [{³√(n²-n³) }+ n ]

Use the formula a³+b³ = (a+b)(a²+b²-ab) in the format 

(a+b) = (a³+b³)/(a²+b²-ab)

here a = ³√(n²-n³)  and b = n

on solving we get 1/(1+1+1) = 1/3