How-to-solve-this-lim-x-1-x-x-2-4x-x-dx-x-2-

Question Number 136278 by adhigenz last updated on 20/Mar/21

How to solve this ?

\lim_{x \to \infty} \frac{\underset{1}{\int^{x}} [\sqrt{x^{2} - 4 x} - x] d x}{x^{2}} = \dots

Answered by MJS_new last updated on 20/Mar/21

error : an integral cannot have dependent

borders . we cannot solve \underset{a}{\int^{x}} f (x) d x

Commented by ajfour last updated on 20/Mar/21

n o e r r o r, i t s h o u l d b e s a m e a s

\int_{a}^{x} f (t) d t .

Answered by mathmax by abdo last updated on 20/Mar/21

i think the Q is \lim_{x \to \infty} \frac{\int_{1}^{x} (\sqrt{t^{2} - 4 t} - t) dt}{x^{2}}

Commented by mathmax by abdo last updated on 20/Mar/21

L =_{hospital} \lim_{x \to \infty} \frac{\sqrt{x^{2} - 4 x} - x}{2 x}

= \lim_{x \to \infty} \frac{(\sqrt{x^{2} - 4 x} - x) (\sqrt{x^{2} - 4 x} + x)}{2 x (\sqrt{x^{2} - 4 x} + x)}

= \lim_{x \to \infty} \frac{x^{2} - 4 x - x^{2}}{2 x (\sqrt{x^{2} - 4 x} + x)} = \lim_{x \to \infty} \frac{- 2}{\sqrt{x^{2} - 4 x} + x} = 0