Question

lim x tends to 0 root 2 - under root 1+ cosx upon sin square x solve pls​

Answer 1

SOLUTION

TO EVALUATE

$\displaystyle \sf{\lim_{x \to 0} \: \frac{ \sqrt{2} - \sqrt{1 + cos x} }{ { \sin}^{2}x }}$

EVALUATION

SOLVE USING L'HOSPITAL RULE

$\displaystyle \sf{\lim_{x \to 0} \: \frac{ \sqrt{2} - \sqrt{1 + cos x} }{ { \sin}^{2}x }} \: \: \: \: \: \: \: \bigg( \: \frac{0}{0} \: \: form \bigg)$

$\displaystyle \sf{ = \lim_{x \to 0} \: \frac{ \frac{ \sin x}{2 \sqrt{1 + cos x}} }{ 2\sin x \cos x}} \:$

$\displaystyle \sf{ = \lim_{x \to 0} \: \frac{ \sin x }{ 4\sin x \cos x \:\sqrt{1 + cos x} }} \:$

$\displaystyle \sf{ = \lim_{x \to 0} \: \frac{ 1 }{ 4 \cos x \:\sqrt{1 + cos x} }} \:$

$\displaystyle \sf{ =\: \frac{ 1 }{ 4 \cos 0 \:\sqrt{1 + cos 0} }} \:$

$\displaystyle \sf{ = \: \frac{ 1 }{ 4 \times 1 \times \:\sqrt{1 + 1} }} \:$

$\displaystyle \sf{ = \: \frac{ 1 }{ 4 \sqrt{2} }} \:$

━━━━━━━━━━━━━━━━

Learn more from Brainly :-

1. the value of limited x to 0 sin7x+sin5x÷tan5x-tan2x is

https://brainly.in/question/28582236

2. Lim x→0 (log 1+8x)/x

https://brainly.in/question/18998465