Algebra> Limx> 0 please see question in images...

Limx> 0 please see question in images

ashu , 11 Years ago

Grade 11

1 Answers

Arun Kumar

$\\\lim_{ x \to \infty }((x^2+1)^{1/2}-(1+x^2)^{1/3})/((x^4+1)^{1/4}-(1+x^4)^{1/5}) \\=\lim_{ x \to \infty }\frac{((x^2+1)^{1/2}-(1+x^2)^{1/3})}{((x^4+1)^{1/4}-(1+x^4)^{1/5})} \\=\frac{\sqrt{x^2+1}-\left(x^2+1\right)^{\frac{1}{3}}}{\left(x^4+1\right)^{\frac{1}{4}}-\left(x^4+1\right)^{\frac{1}{5}}}=\frac{\sqrt{x^2\left(1+\frac{1}{x^2}\right)}-\left(x^2\left(1+\frac{1}{x^2}\right)\right)^{\frac{1}{3}}}{\left(x^4+1\right)^{\frac{1}{4}}-\left(x^4+1\right)^{\frac{1}{5}}} \\=\frac{x\left(\sqrt{1+\frac{1}{x^2}}-\frac{\left(1+\frac{1}{x^2}\right)^{\frac{1}{3}}}{x^{\frac{1}{3}}}\right)}{\left(x^4+1\right)^{\frac{1}{4}}-\left(x^4+1\right)^{\frac{1}{5}}} \\=\frac{x\left(\sqrt{1+\frac{1}{x^2}}-\frac{\left(1+\frac{1}{x^2}\right)^{\frac{1}{3}}}{x^{\frac{1}{3}}}\right)}{\left(x^4\left(1+\frac{1}{x^4}\right)\right)^{\frac{1}{4}}-\left(x^4\left(1+\frac{1}{x^4}\right)\right)^{\frac{1}{5}}}$

$\\=\frac{x\left(\sqrt{1+\frac{1}{x^2}}-\frac{\left(1+\frac{1}{x^2}\right)^{\frac{1}{3}}}{x^{\frac{1}{3}}}\right)}{x\left(\left(1+\frac{1}{x^4}\right)^{\frac{1}{4}}-\frac{\left(1+\frac{1}{x^4}\right)^{\frac{1}{5}}}{x^{\frac{1}{5}}}\right)} \\=\lim _{x\to \infty \:}\left(\frac{x\left(\sqrt{1+\frac{1}{x^2}}-\frac{\left(1+\frac{1}{x^2}\right)^{\frac{1}{3}}}{x^{\frac{1}{3}}}\right)}{x\left(\left(1+\frac{1}{x^4}\right)^{\frac{1}{4}}-\frac{\left(1+\frac{1}{x^4}\right)^{\frac{1}{5}}}{x^{\frac{1}{5}}}\right)}\right) \\=\lim _{x\to \infty \:}\left(\frac{\sqrt{\frac{1}{x^2}+1}-\frac{\left(\frac{1}{x^2}+1\right)^{\frac{1}{3}}}{x^{\frac{1}{3}}}}{\left(\frac{1}{x^4}+1\right)^{\frac{1}{4}}-\frac{\left(\frac{1}{x^4}+1\right)^{\frac{1}{5}}}{x^{\frac{1}{5}}}}\right)$

$\\=\frac{\sqrt{\frac{1}{\infty ^2}+1}-\frac{\left(\frac{1}{\infty ^2}+1\right)^{\frac{1}{3}}}{\infty ^{\frac{1}{3}}}}{\left(\frac{1}{\infty ^4}+1\right)^{\frac{1}{4}}-\frac{\left(\frac{1}{\infty ^4}+1\right)^{\frac{1}{5}}}{\infty ^{\frac{1}{5}}}} \\=1$