IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

anandghegde

for the second question,
use $a-b=(a^{\frac{1}{3}} -b^{\frac{1}{3}})(a^{\frac{2}{3}}+b^{\frac{2}{3}}+(ab)^{\frac{1}{3}})$

here a= x^3+x^2+1

and

.

$a^{\frac{1}{3}}-b^{\frac{1}{3}} = \frac{a-b}{a^{\frac{2}{3}}+b^{\frac{2}{3}}+(ab)^{\frac{1}{3}}}$

Hence

$lim_{x$

= $\displaystyle \lim_{x\to\infty} \frac{x^2+1}{(x^3+x^2+1)^{\frac{2}{3}}+x^2+x.(x^3+x^2+1)^{\frac{1}{3}}}$

divide numerator and denominator by x^2, we get

$\displaystyle \lim_{x\to\infty} \frac{1+\frac{1}{x^2}}{(1+\frac{1}{x}+\frac{1}{x^2})^{\frac{2}{3}}+1+.(1+\frac{1}{x}+\frac{1}{x^2})^{\frac{1}{3}}}$

$=\frac{1}{1+1+1} =\frac{1}{3}$

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