let-give-F-x-x-2x-dt-1-t-2-t-4-1-calculate-dF-dx-x-2-find-lim-x-F-x-and-lim-x-F-x-x-

Question Number 29456 by prof Abdo imad last updated on 08/Feb/18

l e t g i v e F (x) = \int_{x}^{2 x} \frac{d t}{\sqrt{1 + t^{2} + t^{4}}}

1) c a l c u l a t e \frac{d F}{d x} (x)

2) f i n d {l i m}_{x \to + \infty} F (x) a n d {l i m}_{x \to + \infty} \frac{F (x)}{x} .

Commented by prof Abdo imad last updated on 13/Feb/18

1) F^{^{'}} (x) = \frac{2}{\sqrt{1 + 4 x^{2} + 16 x^{4}}} - \frac{1}{\sqrt{1 + x^{2} + x^{4}}} .

2) w e h a v e 1 + t^{2} + t^{4} > t^{2} + t^{4} \Rightarrow \sqrt{1 + t^{2} + t^{4} >} t \sqrt{1 + t^{2}}

f o r t > 0 \Rightarrow \frac{1}{\sqrt{1 + t^{2} + t^{4}}} < \frac{1}{t \sqrt{1 + t^{2}}} \Rightarrow

F (x) < \int_{x}^{2 x} \frac{d t}{t \sqrt{1 + t^{2}}} < \int_{x}^{2 x} \frac{d t}{t^{2}} = {[- \frac{1}{t}]}_{x}^{2 x} = \frac{1}{x} - \frac{1}{2 x}

b u t {l i m}_{x \to + \infty} (\frac{1}{x} - \frac{1}{2 x}) = 0 \Rightarrow {l i m}_{x \to + \infty} F (x) = 0

a l s o {l i m}_{x \to + \infty} \frac{F (x)}{x} = 0 .