Squeeze Theorem

We don’t prove any derivatives in this post, but we need these results in order to evaluate the trigonometric derivatives in the next part. We also get to showcase a very useful theorem in calculus.


The Statement of the Squeeze Theorem

Let $a \leq c \leq b$. Suppose $g(x) \leq f(x) \leq h(x)$ for all $x \in (a, c) \cup (c, b)$. Then

\[\lim_{x \rightarrow c} g(x) \leq \lim_{x \rightarrow c} f(x) \leq \lim_{x \rightarrow c} h(x)\]


A corollary to the theorem is that if $\displaystyle \lim_{x \rightarrow c} g(x) = \lim_{x \rightarrow c} h(x) = L$, the $\displaystyle \lim_{x \rightarrow c} f(x) = L$.

Intuitively, $g(x)$ serves as a lower-bound and $h(x)$ an upper-bound for $f(x)$ in the neighborhood around $c$. Thus, these bounds are preserved in the limit. Thus, if the limits are equal, the upper and lower-bound squeeze the limit of $f$ to a single number.


Applying the Squeeze Theorem

First, we use the squeeze theorem to compute $\displaystyle \lim_{x \rightarrow 0} \frac{\sin x}{x}$. Notice that direct substitution yields $\frac{0}{0}$, which does not give us any information on the actual value of the limit.

The proof hinges on the inequality, $\displaystyle \cos x \leq \frac{\sin x}{x} \leq 1$ for all $x \in (-\frac{\pi}{2}, 0) \cup (0, \frac{\pi}{2})$. Then, we just apply the squeeze theorem to see that

\[\begin{align} &\lim_{x \rightarrow 0} \cos x &\leq& \quad \lim_{x \rightarrow 0} \frac{\sin x}{x} &\leq& \quad \lim_{x \rightarrow 0} 1 \\[10pt] &1 &\leq& \quad \lim_{x \rightarrow 0} \frac{\sin x}{x} &\leq& \quad 1 \end{align}\]

Therefore, $\displaystyle \lim_{x \rightarrow 0} \frac{\sin x}{x} = 1$.

Now all that is left is to prove the original inequality. Consider the following construction on a unit circle. From this, we obtain three areas. Please refer to this post for how the heights of the top and bottom triangles are derived. Hint: use the definitions of the trigonometric functions and similar triangles.

    


Label the three areas $A_1$, $A_2$, and $A_3$ from bottom to top. We apply the appropriate area formulas to see that

\[\begin{align} &A_1 &\quad=\quad& \frac{1}{2}bh &\quad=\quad& \frac{1}{2} \cdot 1 \cdot \sin x &\quad=\quad& \frac{1}{2} \sin x \\[10pt] &A_2 &\quad=\quad& \frac{1}{2} r^2 \theta &\quad=\quad& \frac{1}{2} \cdot 1^2 \cdot x &\quad=\quad& \frac{1}{2} x\\[10pt] &A_3 &\quad=\quad& \frac{1}{2}bh &\quad=\quad& \frac{1}{2} \cdot 1 \cdot \tan x &\quad=\quad& \frac{1}{2} \tan x \end{align}\]

Clearly, $A_1 \leq A_2 \leq A_3$, since each area can be contained inside the other. Therefore,

\[\begin{align} &A_1 \leq A_2 \leq A_3 \\[10pt] &\frac{1}{2} \sin x \leq \frac{1}{2}x \leq \frac{1}{2} \tan x \\[10pt] &\sin x \leq x \leq \tan x \end{align}\]

Now, this is only true on the interval, $(0, \frac{\pi}{2})$, since outside this interval, our geometric argument breaks down. Observe that $\sin x$, $x$, and $\tan x$ are all odd functions. Recall that any odd function $f$ has the property that $f(-x) = -f(x)$. Therefore,

\[\begin{align} &\sin x \leq x \leq \tan x\\[10pt] &-\sin x \geq -x \geq -\tan x\\[10pt] &\sin (-x) \geq -x \geq \tan (-x) \\ \end{align}\]

Thus, we get the following result

\[\begin{cases} \sin x \leq x \leq \tan x &\qquad\text{if } 0 < x < \frac{\pi}{2} \\ \sin x \geq x \geq \tan x &\qquad\text{if } -\frac{\pi}{2} < x < 0 \end{cases}\]

Divide all parts of the equations by $\sin(x)$. Notice, that in the case where $-\frac{\pi}{2} < x < 0$, this value is always negative and thus the inequality will flip. Thus, both cases will be the same, and we get this result for all $x \in (-\frac{\pi}{2}, 0) \cup (0, \frac{\pi}{2})$.

\[\begin{align} &1 \leq \frac{x}{\sin x} \leq \frac{1}{\cos x}\\[10pt] &1 \geq \frac{\sin x}{x} \geq \cos x \end{align}\]

Which is the result we needed to prove. Below is a graphical representation of this proof. We can see how the light blue line is bounded by the other two lines in the neighborhood around $0$.

   


Corollary

As a corollary, we can compute $\displaystyle \lim_{x \rightarrow 0} \frac{1 - \cos x}{x}$. Notice that direct substitution yields $\frac{0}{0}$, which does not give us any information on the actual value of the limit.

\[\begin{align} \lim_{x \rightarrow 0} \frac{1 - \cos x}{x} &= \lim_{x \rightarrow 0} \frac{1 - \cos x}{x} \cdot \frac{1 + \cos x}{1 + \cos x} \\[10pt] &= \lim_{x \rightarrow 0} \frac{1 - \cos^2 x}{x(1 + \cos x)} \\[10pt] &= \lim_{x \rightarrow 0} \frac{\sin^2 x}{x(1 + \cos x)} \\[10pt] &= \left ( \lim_{x \rightarrow 0} \frac{\sin x}{x} \right ) \left ( \lim_{x \rightarrow 0} \frac{\sin x}{1 + \cos x} \right ) \\[10pt] &= 1 \cdot \frac{0}{1+1} \\[10pt] &= 0 \end{align}\]

Derivative Proofs Series