7. Computing Limits

c. Sandwich Theorem

2. Sine and Cosine Limits

The most important application of the Sandwich Theorem is to prove the following four limits about \(\sin\) and \(\cos\). \[\begin{aligned} \lim_{\theta\to0}\sin\theta&=0 &\qquad&\lim_{\theta\to0}\cos\theta=1\\ \lim_{\theta\to0}\dfrac{\sin\theta}{\theta}&=1 &\qquad&\lim_{\theta\to0}\dfrac{1-\cos\theta}{\theta}=0 \end{aligned}\] The first two seem obvious since \[ \sin0=0 \qquad \cos0=1 \] However, in principle, the value of a function may or may not be the same as the limit. The formal proofs involve the Sandwich Theorem. (They do not need the precise definition of limits.)

\[ \lim_{\theta\to0}\sin\theta=0 \qquad\lim_{\theta\to0}\cos\theta=1 \]


Let's first assume \(\theta\gt0\). The diagram shows a unit circle and a ray at an angle \(\theta\) intersecting the circle at the point \(B=(\cos\theta,\sin\theta)\). The circumference of the full circle is \(2\pi r=2\pi\). The fraction of the circumference occupied by the arc \(\stackrel{\frown}{AB}\) is \(\dfrac{\theta}{2\pi}\) (in radians). So the length of the arc is \(s=\dfrac{\theta}{2\pi}2\pi=\theta\). The perpendicular from \(B\) to the \(x\)-axis has length \(y=\sin\theta\) and since the length of a perpendicular is shortest distance, we conclude \[ 0 \le \sin\theta \le s = \theta \] So \(\sin\theta\) is squeezed between \(0\) and \(\theta\). Since \[ \lim_{\theta\to0^+}0=\lim_{\theta\to0^+}\theta=0 \] we conclude \(\displaystyle \lim_{\theta\to0^+}\sin\theta=0\) also.

prop_SinLim

Now, let's assume \(\theta\lt0\). We let \(t=-\theta\) and use the fact that \(\sin\) is odd, to compute: \[ \lim_{\theta\to0^-}\sin\theta =\lim_{t\to0^+}\sin(-t) =-\lim_{t\to0^+}\sin(t) =0 \]



This limit does not need the Sandwich Theorem. We just use a trig identity the Limit Laws and the previous limit. \[\begin{aligned} \lim_{\theta\to0}\cos\theta &=\lim_{\theta\to0}\sqrt{1-\sin^2\theta} \\ &=\sqrt{\lim_{\theta\to0}\left(1-\sin^2\theta\right)} \\ &=\sqrt{1-\left(\lim_{\theta\to0}\sin\theta\right)^2} \\ &=\sqrt{1-(0)^2}=1 \\ \end{aligned}\]

\[ \lim_{\theta\to0}\dfrac{\sin\theta}{\theta}=1 \qquad\lim_{\theta\to0}\dfrac{1-\cos\theta}{\theta}=0 \]


The diagram shows a unit circle and a ray at an angle \(\theta\) intersecting the circle at the point \(B=(\cos\theta,\sin\theta)\). The area of the full circle is \(\text{Area}_\bigcirc=\pi r^2=\pi\). The fraction of the area occupied by the sector \(\measuredangle OAB\) is \(\dfrac{\theta}{2\pi}\) (in radians). So the area of the sector is \[ \text{Area}_{\measuredangle OAB}=\dfrac{\theta}{2\pi}\pi=\dfrac{\theta}{2} \] This sector is sandwiched between two triangles. The smaller triangle is \(\triangle OAB\) whose area is \[ \text{Area}_{\triangle OAB}=\dfrac{1}{2}(1)\sin\theta \] The larger triangle is \(\triangle OAC\) whose area is \[ \text{Area}_{\triangle OAC}=\dfrac{1}{2}(1)\tan\theta \]

prop_SinQLim

The three areas are related by: \[ \text{Area}_{\triangle OAB} \le \text{Area}_{\measuredangle OAB} \le \text{Area}_{\triangle OAC} \] \[ \dfrac{1}{2}\sin\theta \le \dfrac{\theta}{2} \le \dfrac{1}{2}\tan\theta \] The first inequality can be rearranged to say: \[ \dfrac{\sin\theta}{\theta} \le 1 \] The second inequality can be rearranged to say: \[ 1 \le \dfrac{\tan\theta}{\theta}=\dfrac{\sin\theta}{\theta\cos\theta} \] or \[ \cos\theta \le \dfrac{\sin\theta}{\theta} \] Together, they say \(\dfrac{\sin\theta}{\theta}\) is squeezed between \(\cos\theta\) and \(1\). Since \[ \lim_{\theta\to0}\cos\theta=\lim_{\theta\to0}1=1 \] we conclude \(\displaystyle \lim_{\theta\to0}\dfrac{\sin\theta}{\theta}=1\) also.



This limit does not need the Sandwich Theorem. We multiply by the conjugate, use a trig identity and the previous limits. \[\begin{aligned} \lim_{\theta\to0}\dfrac{1-\cos\theta}{\theta} &=\lim_{\theta\to0}\dfrac{(1-\cos\theta)}{\theta}\dfrac{(1+\cos\theta)}{(1+\cos\theta)} \\ &=\lim_{\theta\to0}\dfrac{(1-\cos^2\theta)}{\theta^2}\dfrac{\theta}{(1+\cos\theta)} \\ &=\lim_{\theta\to0}\dfrac{\sin^2\theta}{\theta^2}\dfrac{\theta}{(1+\cos\theta)} \\ &=\left(\lim_{\theta\to0}\dfrac{\sin\theta}{\theta}\right)^2 \lim_{\theta\to0}\dfrac{\theta}{1+\cos\theta} \\ &=(1)^2\left(\dfrac{0}{1+1}\right)=0 \end{aligned}\]

Compute \(\displaystyle \lim_{x\to0}\dfrac{2x+3x^2}{\sin x}\)

Divide the numerator and denominator by \(x\). Then use the Quotient Law and the limit \(\displaystyle \lim_{x\to0}\dfrac{\sin x}{x}=1\): (Note the variable does not matter.) \[\begin{aligned} \lim_{x\to0}&\dfrac{2x+3x^2}{\sin x} =\lim_{x\to0}\dfrac{2+3x}{\dfrac{\sin x}{x}} \\ &=\dfrac{\displaystyle \lim_{x\to0}(2+3x)}{\displaystyle \lim_{x\to0}\dfrac{\sin x}{x}} =\dfrac{2}{1}=2 \end{aligned}\]

Compute \(\displaystyle \lim_{\theta\to0}\dfrac{1-\cos\theta}{\theta^2}\).

Multiply numerator and denominator by the conjugate and use a trig identity.

\(\displaystyle \lim_{\theta\to0}\dfrac{1-\cos\theta}{\theta^2}=\dfrac{1}{2}\)

We multiply by the conjugate, use a trig identity and the previous limits. \[\begin{aligned} \lim_{\theta\to0}\dfrac{1-\cos\theta}{\theta^2} &=\lim_{\theta\to0}\dfrac{(1-\cos\theta)}{\theta^2}\dfrac{(1+\cos\theta)}{(1+\cos\theta)} \\ &=\lim_{\theta\to0}\dfrac{(1-\cos^2\theta)}{\theta^2}\dfrac{1}{(1+\cos\theta)} \\ &=\lim_{\theta\to0}\dfrac{\sin^2\theta}{\theta^2}\dfrac{1}{(1+\cos\theta)} \\ &=\left(\lim_{\theta\to0}\dfrac{\sin\theta}{\theta}\right)^2 \lim_{\theta\to0}\dfrac{1}{1+\cos\theta} \\ &=(1)^2\left(\dfrac{1}{1+1}\right)=\dfrac{1}{2} \end{aligned}\]

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