limit(sqrt(x^2+1)-sqrt(x^2-4x)), x approaches infinity

Question

mathmath333 · Answer

$\large\tt \color{black}{\lim_{x\to\infty}\sqrt{(x^2+1)}-\sqrt{(x^2-4x)}}$

$\large\tt \color{black}{\lim_{x\to\infty}\sqrt{(x^2+1)}-\sqrt{(x^2-4x)} \times \dfrac{\sqrt{(x^2+1)}+\sqrt{(x^2-4x)}}{\sqrt{(x^2+1)}+\sqrt{(x^2-4x)}}} $

$\large\tt \color{black}{\lim_{x\to\infty} \dfrac{(x^2+1)-(x^2-4x)}{\sqrt{(x^2+1)}+\sqrt{(x^2-4x)}}} $

$\large\tt \color{black}{\lim_{x\to\infty} \dfrac{1+4x}{\sqrt{(x^2+1)}+\sqrt{(x^2-4x)}}} $

$\large\tt \color{black}{\lim_{x\to\infty} \dfrac{x(\frac{1}{x}+4)}{x\sqrt{(1+\frac{1}{x^2})}+\sqrt{(1-\frac{4}{x})}}} $

$\large\tt \color{black}{\lim_{x\to\infty} \dfrac{\frac{1}{x}+4)}{\sqrt{(1+\frac{1}{x^2})}+\sqrt{(1-\frac{4}{x})}}} $

$\large\tt \color{black}{ =\dfrac{(\frac{1}{\infty}+4)}{\sqrt{(1+\frac{1}{\infty^2})}+\sqrt{(1-\frac{4}{\infty})}}} $

$\large\tt \color{black}{ =\dfrac{(0+4)}{\sqrt{(1+0)}+\sqrt{(1-0)}}} $

$\large\tt \color{black}{=2}$