limit

is it 1 ?

=(2010)²⁰¹²

Yes man111. Your answer is correct.

We have,

L = \lim_{n\to\infty} \left(2012\sqrt[n]{2010}-2011 \right)^n

L = \lim_{n\to\infty} \left(1+2012(\sqrt[n]{2010}-1))^n

It is of the form 1^∞.

L = e^{ \lim_{n\to\infty} 2012(\sqrt[n]{2010}-1\;)\;n}

Substituting n = 1/r ; lim _{n→ ∞} = lim _{r → 0}

L = e^{ \lim_{r\to0} 2012(2010^r-1\;)\;\frac{1}{r}}

By L-Hopitals in the index limit,

L = e^{2012.\ln{2010}}

L = 2010^{2012} \;\; \blacksquare \mathcal{H}\text{ence}\;\mathcal{D}\text{one}

- Vivek