Limits

Calculus Level 3

Evaluate the limit

lim ⁡ θ → 0 sin ⁡ θ sin ⁡ 2 θ 1 − cos ⁡ θ \large \lim_{\theta \rightarrow 0} \dfrac{\sin \theta \sin 2\theta}{1-\cos \theta}

2 3 4 0

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1 solution

lim ⁡ θ → 0 sin ⁡ θ sin ⁡ 2 θ 1 − cos ⁡ θ \large \displaystyle \lim_{\theta \rightarrow 0} \dfrac{\sin\theta\sin2\theta}{1-\cos\theta}

= lim ⁡ θ → 0 sin ⁡ θ ( 2 ) ( sin ⁡ θ ) ( cos ⁡ θ ) 1 − cos ⁡ θ \large \displaystyle=\lim_{\theta \to 0} \dfrac{\sin\theta(2)(\sin\theta)(\cos\theta)}{1-\cos\theta}

= lim ⁡ θ → 0 2 sin ⁡ 2 cos ⁡ θ 1 − cos ⁡ θ \large \displaystyle=\lim_{\theta \to 0} \dfrac{2\sin^2\cos\theta}{1-\cos\theta}

= lim ⁡ θ → 0 2 ( 1 − cos ⁡ 2 θ ) ( cos ⁡ θ ) 1 − cos ⁡ θ \large \displaystyle=\lim_{\theta \to 0} \dfrac{2(1-\cos^2\theta)(\cos\theta)}{1-\cos\theta}

= lim ⁡ θ → 0 2 ( 1 − cos ⁡ θ ) ( 1 + cos ⁡ θ ) ( cos ⁡ θ ) 1 − cos ⁡ θ \large \displaystyle=\lim_{\theta \to 0} \dfrac{2(1-\cos\theta)(1+\cos\theta)~(\cos\theta)}{1-\cos\theta}

= lim ⁡ θ → 0 ( 2 ) ( 1 + cos ⁡ θ ) ( cos ⁡ θ ) \large \displaystyle=\lim_{\theta \to 0} {(2)(1+\cos\theta)(\cos\theta)}

= 2 ( 1 + 1 ) ( 1 ) \large=2(1+1)(1)

= 2 ( 2 ) \large=2(2)

= 4 \large=4

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Nice solution. One minor typo on line 2.

Marta Reece - 4 years ago

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