Question

Medal + Fan

Answer 1

\[\lim_{x \rightarrow 0} \frac{ x \sin(x) }{ 1-\cos(x) }\]

Answer 2

\[1-\cos x=2\sin ^2\frac{ x }{ 2 }\]

Answer 3

\[\huge \lim_{x \rightarrow 0} \frac {x \sin x}{1-\cos x}\]
\[\huge \lim_{x \rightarrow 0} \frac {x \sin x}{1-\cos x} \times \frac {(1+\cos x)}{(1+\cos x)}\]
\[= \huge \lim_{x \rightarrow 0} \frac {x \sin x(1+\cos x)}{1-\cos^2x} \]
\[= \huge \lim_{x \rightarrow 0} \frac {x \sin x(1+\cos x)}{\sin^2x} \]
\[= \huge \lim_{x \rightarrow 0} \frac {x (1+\cos x)}{sinx} \]
\[= \huge \lim_{x \rightarrow 0} \frac {x }{sinx} \times \lim_{x \rightarrow 0} (1+\cos x)\]
\[= \huge \lim_{x \rightarrow 0} \frac {\frac{x}{x} }{\frac{sinx}{x}} \times \lim_{x \rightarrow 0} (1+\cos x)\]
\[= \huge \lim_{x \rightarrow 0} \frac {1 }{\frac{sinx}{x}} \times \lim_{x \rightarrow 0} (1+\cos x)\]
\[= \huge \lim_{x \rightarrow 0} \frac {1 }{\frac{sinx}{x}} \times \lim_{x \rightarrow 0} (1+\cos x)\]
\[= \huge \frac {1 }{1} \times (1+\cos0^0) = 1+ 1 = 2\]

@HemerickLee

Answer 4

\[\lim_{x \rightarrow 0}\frac{ xsinx }{ 2\sin ^2\frac{ x }{ 2 } }=\lim_{x \rightarrow 0}\frac{ \frac{ x \sin x }{ x^2 } }{ 2\frac{ \sin ^2\frac{ x }{ 2 } }{ \frac{ x^2 }{ 4 }*4 } }\]
\[=2\frac{ \lim_{x \rightarrow 0}\frac{ \sin x }{ x } }{ \lim_{x \rightarrow 0}\left( \frac{ \sin \frac{ x }{ 2 } }{ \frac{ x }{ 2 } } \right)^2 }\]
\[=2\times \frac{ 1 }{ 1^2 }=2\]