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Question Number 42808 by maxmathsup by imad last updated on 02/Sep/18

l e t f (x) = \int_{x}^{2 x} \frac{1 + c o s (t)}{\sqrt{t^{4} - t^{2} + 4}} d t

1) f i n d D_{f}

2) c a l c u l a t e f^{^{'}} (x)

3) f i n d {l i m}_{x \to + \infty} f (x)

Answered by tanmay.chaudhury50@gmail.com last updated on 04/Sep/18

f^{'} (x) = \int_{x}^{2 x} \frac{\partial}{\partial x} (\frac{1 + c o s t}{\sqrt{t^{4} - t^{2} + 4}}) d x + \frac{1 + c o s 2 x}{\sqrt{{(2 x)}^{4} - {(2 x)}^{2} + 4}} \times \frac{d (2 x)}{d x} - \frac{1 + c o s x}{\sqrt{x^{4} - x^{2} + 4}} \times \frac{d (x)}{d x}

= 2 \times \frac{1 + c o s 2 x}{\sqrt{16 x^{4} - 4 x^{2} + 4}} - \frac{1 + c o s x}{\sqrt{x^{4} - x^{2} + 4}}

f o r a n y v a l u e o f x t h e v a l u e 1 ⩾ c o s x ⩾ - 1

a n d 1 ⩾ c o s 2 x ⩾ - 1

s o

\lim_{x \to \infty} \frac{2 + 2 c o s 2 x}{\sqrt{16 x^{4} - 4 x^{2} + 4}} - \frac{1 + c o s x}{\sqrt{x^{4} - x^{2} + 4}}

\lim_{x \to \infty} \frac{p (x)}{q (x)} - \frac{g (x)}{h (x)}

w h e n x \to \infty p (x) \to 4 o r 0 b u t q (x) \to \infty

w h e n x \to \infty g (x) \to 2 o r 0 b u t h (x) \to \infty

s o \lim_{x \to \infty} \frac{p (x)}{q (x)} - \frac{g (x)}{h (x)} \to 0