Question

please give me a valid answer
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Answer 1

$\sf lim_{x \rightarrow 0} \: \dfrac{1 - cosx}{xsinx}$

• If we do direct substitution then resultant value is in the form of 0/0

$\sf lim_{x \rightarrow 0} \: \dfrac{1 -cosx}{xsinx}$

• Multiply by (1-cosx)/(1-cosx)

$\sf lim_{x \rightarrow 0} \: \dfrac{1 - cosx}{xsinx} \times \dfrac{1 + cosx}{1 + cosx}$

$\sf lim_{x \rightarrow 0} \: \dfrac{1 - cos^{2} x}{xsinx} \times \dfrac{1}{1 + cosx}$

• Trigonometric identity = cos²x + sin²x = 1
• 1 - cos²x = sin²x

$\sf lim_{x \rightarrow 0} \: \dfrac{sin^{2} x}{xsinx} \times \dfrac{1}{1 + cosx}$

$\sf lim_{x \rightarrow 0} \: \dfrac{sin x \times \cancel {sinx}}{x \times \cancel{sinx}} \times \dfrac{1 }{1 + cosx}$

$\sf lim_{x \rightarrow 0} \: \dfrac{sin x}{x} \times \dfrac{1 }{1 + cosx}$

• sinx/x = 1
• Substitute the value of x = 0

$\sf lim_{x \rightarrow 0} \: 1\times \dfrac{1 }{1 + cosx} = \dfrac{1}{1 + 1} = \dfrac{1}{2}$

$\qquad {\boxed{\bf lim_{x \rightarrow 0} \: \dfrac{1 - cosx}{xsinx}=\dfrac{1}{2}}}$

Answer 2

Answer:

REQUIRED ANSWER:

$\sf{lim \: ˣ \rightarrow0 \frac{1 - cos \: x}{x \:sin \: x} }$

$\sf{limˣ \rightarrow0 \frac{1 - \cos \: x }{x \: sin \: x} \times \frac{1 + cos \: x}{1 + cos \: x} }$

$\sf{limˣ \rightarrow0 \frac{1 - {cos}^{2} }{x \: sin \: x } \times \frac{1}{1 + \: cos \: x} }$

$\sf{limˣ \rightarrow0 \frac{ { \sin }^{2}x }{ \: x \: sin \: x} \times \frac{1}{1 + cos \: x} }$

$\sf{limˣ \rightarrow0 \frac{sin \: x}{x} \times \frac{1}{1 + cos \: x} }$

$\sf{limˣ \rightarrow0 \: 1 \times \frac{1}{1 + cos \: x} = \frac{1}{1 + 1} = \frac{1}{2} }$

$\red \bigstar\boxed{limˣ \rightarrow0 \: \frac{1 - cosx}{xsinx} = \frac{1}{2} }$