X-lim-x-sin x-x-cos x- -does not exist

The correct option is D x→∞lim√(x+sinx/x cosx) =x→∞limx^1/2√(1+sinx/x)/x^1/2√(1 cosx/x) We know, that for any value of x, sinx and cosx will be [ 1,1] So, =x→ ...

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Byju's Answer

Standard XI

Mathematics

Sandwich Theorem

x→∞lim√(x+sin...

Question

$lim x \to \infty \sqrt{\frac{x + sin x}{x - cos x}} =$

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does not exist

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Solution

The correct option is B
$lim x \to \infty \sqrt{\frac{x + s i n x}{x - c o s x}}$
$= lim x \to \infty \frac{x^{\frac{1}{2}} \sqrt{1 + \frac{s i n x}{x}}}{x^{\frac{1}{2}} \sqrt{1 - \frac{c o s x}{x}}}$
We know, that for any value of x, sinx and cosx will be [-1,1]
So, $= lim x \to \infty \frac{S i n x}{x} = 0$
And $= lim x \to \infty \frac{C o s x}{x} = 0$
$= x^{\frac{1}{2}} \sqrt{1 + \frac{s i n x}{x}} \frac{\sqrt{1 + \frac{s i n x}{x}}}{\sqrt{1 + \frac{c o s x}{x}}}$
$= lim x \to \infty \frac{1}{1}$
$= 1$

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limx→∞√x+sinxx−cosx=

The correct option is B limx→∞√x+sinxx−cosx =limx→∞x12√1+sinxxx12√1−cosxx We know, that for any value of x, sinx and cosx will be [-1,1] So, =limx→∞Sinxx=0 And =limx→∞Cosxx=0 =x12√1+sinxx√1+sinxx√1+cosxx =limx→∞11 =1

$lim x \to \infty \sqrt{\frac{x + sin x}{x - cos x}} =$