Evaluate 1 lim x tends to zero root(2)--root(1+cosx) /sin square x - Maths - Limits and Derivatives

Question

Evaluate 1 lim x tends to zero root(2)--root(1+cosx) /sin square
x
2 lim x tends to pai/2 (cos 3x+3cosx) / ( pai/2-x)whole cube
3 lim x tends to 2 (x/x-2 --- 4/xsqare - 2x)

Ankush Jain · Accepted Answer

1)

If your query is: $\lim_{x \to 0} \frac{\sqrt{2} - \sqrt{1 + \cos x}}{x^{2}}$ then the solution follows as:

We have,

$\lim_{x \to 0} \frac{\sqrt{2} - \sqrt{1 + \cos x}}{x^{2}} = \lim_{x \to 0} \frac{\sqrt{2} - \sqrt{2 \cos^{2} \frac{x}{2}}}{x^{2}} = \lim_{x \to 0} \frac{\sqrt{2} (1 - \cos \frac{x}{2})}{x^{2}} = \lim_{x \to 0} \frac{\sqrt{2} (2 \sin^{2} \frac{x}{4})}{x^{2}} [u \sin g h a l f a n g l e f o r m u l a f o r 1 - \cos \frac{x}{2}] = \lim_{x \to 0} \frac{2 \sqrt{2} \sin^{2} \frac{x}{4}}{{(\frac{x}{4})}^{2} \times 4^{2}} = \frac{2 \sqrt{2}}{4^{2}} \lim_{x \to 0} {(\frac{\sin \frac{x}{4}}{\frac{x}{4}})}^{2} = \frac{2 \sqrt{2}}{4 \times 4} \times 1^{2} = \frac{\sqrt{2}}{8}$

2)
We have,

$\lim_{x \to \frac{π}{2}} \frac{\cos 3 x + 3 \cos x}{(\frac{π}{2} - x)^{3}} = \lim_{x \to \frac{π}{2}} \frac{4 \cos^{3} x - 3 \cos x + 3 \cos x}{(\frac{π}{2} - x)^{3}} [u \sin g : \cos 3 x = 4 \cos^{3} x - 3 \cos x] = \lim_{x \to \frac{π}{2}} \frac{4 \cos^{3} x}{(\frac{π}{2} - x)^{3}} = \lim_{x \to \frac{π}{2}} \frac{4 \sin^{3} (\frac{π}{2} - x)}{(\frac{π}{2} - x)^{3}} = 4 \lim_{x \to \frac{π}{2}} {[\frac{\sin (\frac{π}{2} - x)}{(\frac{π}{2} - x)}]}^{3} = 4 \times 1^{3} = 4$

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Evaluate 1. lim x tends to zero root(2)--root(1+cosx) /sin square x 2. lim x tends to pai/2 (cos 3x+3cosx) / ( pai/2-x)whole cube 3.lim x tends to 2 (x/x-2 --- 4/xsqare - 2x)

Evaluate 1. lim x tends to zero root(2)--root(1+cosx) /sin square x

2. lim x tends to pai/2 (cos 3x+3cosx) / ( pai/2-x)whole cube

3.lim x tends to 2 (x/x-2 --- 4/xsqare - 2x)