Find the limit of xsin⁡(x) as x approaches 0.

Let f(x)=xsin⁡(x)=f′(x)=limx⇒0xsin⁡(x)=f′(x)=limx⇒01sin⁡(x)x=limx⇒01limx⇒0(sin⁡(x)x)=11=1

limx⇒0xsin⁡(x)=limx⇒0(sin⁡(x)x)−1=(limx⇒0sin⁡(x)x)−1=1−1=1

Let f(x)=xsin⁡(x)=f′(x)=limx⇒0xsin⁡(x)=f′(x)=limx⇒01sin⁡(x)x=limx⇒01limx⇒0(sin⁡(x)x)=11=1

limx⇒0xsin⁡(x)=limx⇒0(sin⁡(x)x)−1=(limx⇒0sin⁡(x)x)−1=1−1=1

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lunnatican4

2021-12-18

Find the limit of

\frac{x}{\sin (x)}

as x approaches 0.

Kirsten Davis

Beginner2021-12-19Added 27 answers

Let

f (x) = \frac{x}{\sin (x)}

= f^{'} (x) = lim_{x \Rightarrow 0} \frac{x}{\sin (x)}

= f^{'} (x) = lim_{x \Rightarrow 0} \frac{1}{\frac{\sin (x)}{x}} = \frac{lim_{x \Rightarrow 0} 1}{lim_{x \Rightarrow 0} (\frac{\sin (x)}{x})} = \frac{1}{1} = 1

Cassandra Ramirez

Beginner2021-12-20Added 30 answers

lim_{x \Rightarrow 0} \frac{x}{\sin (x)} = lim_{x \Rightarrow 0} {(\frac{\sin (x)}{x})}^{- 1} = {(lim_{x \Rightarrow 0} \frac{\sin (x)}{x})}^{- 1} = 1^{- 1} = 1

RizerMix

Expert2021-12-29Added 656 answers

$lim_{x \Rightarrow 0} \frac{x}{\sin (x)} = 1$

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