Applications of continuous functions

## Applications of continuous functions
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            ## Classes of continuous functions
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Constant functions

Linear functions

Power functions

Polynomial functions

Rational functions

Exponential functions

Logarithmic functions

Trigonometric functions

Inverse trigonometric functions

Absolute value function
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and, for $f,g$ continuous and $b,c$ constant:

$bf+cg$

$f\times g$

$f/g$ except at zeros of $g$

$f \circ g$ provided<br>
              $\mathrm{range}(g) \subseteq \mathrm{domain}(f)$
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            ---
            ## One reason to care about continuity
            #### Example
            Evaluate $\displaystyle \lim\_{x\to3} f(x)$, where
            $\displaystyle
              f(x) = \sin\left( \frac{3x-2}{7x^2-\log(5x)} \right)
            $.
            
            ---
            ## One reason to care about continuity
            #### Example
            Evaluate $\displaystyle \lim\_{x\to3} f(x)$, where
            $\displaystyle
              f(x) = \sin\left( \frac{3x-2}{7x^2-\log(5x)} \right)
            $.
            #### Solution
            The function $f$ is continuous at $3$, so

$\displaystyle \lim\_{x\to3} f(x)=f(3) = \sin\left( \frac{7}{63-\log(15)} \right)$.
            
            ---
            ## One reason to care about continuity
            #### Example
            Evaluate $\displaystyle \lim\_{x\to3} f(x)$, where
            $\displaystyle
              f(x) = \sin\left( \frac{3x-2}{7x^2-\log(5x)} \right)
            $.
            #### Solution
            The function $f$ is continuous at $3$, so

$\displaystyle \lim\_{x\to3} f(x)=f(3) = \sin\left( \frac{7}{63-\log(15)} \right)$.
            #### Check this works
            Carefully check that $f$ is continuous at $3$ using the classes of continuous functions.
            
            ---
            ## Using continuity to evaluate limits
            #### Example
            Evaluate $\displaystyle \lim\_{x\to0} f(x)$, $\displaystyle \lim\_{x\to0} g(x)$, $\displaystyle \lim\_{x\to\frac\pi2} q(x)$, where
            $\displaystyle
              f(x) = \frac{4x}{x+1}\cos\left( \frac1x \right)
            $,

$\displaystyle
              g(x) = ([\operatorname{sgn}(x)]^2+5x)^2
            $,  
            $\displaystyle
              q(x) = \sin([\tan(x)]^2)
            $.
            
            ---
            ## Using continuity to evaluate limits
            #### Example
            Evaluate $\displaystyle \lim\_{x\to0} f(x)$, $\displaystyle \lim\_{x\to0} g(x)$, $\displaystyle \lim\_{x\to\frac\pi2} q(x)$, where
            $\displaystyle
              f(x) = \frac{4x}{x+1}\cos\left( \frac1x \right)
            $,

$\displaystyle
              g(x) = ([\operatorname{sgn}(x)]^2+5x)^2
            $,  
            $\displaystyle
              q(x) = \sin([\tan(x)]^2)
            $.

#### Solution
            We cannot evaluate $\displaystyle \lim\_{x\to0} f(x)$ using continuity, because $0$ is not in the domain of $\frac1x$, so $f$ is not continuous at $0$.

But we can distribute limit over the product as

$\displaystyle \lim\_{x\to0} f(x) = \left(\lim\_{x\to0} \frac{4}{1+x}\right) \times \left(\lim\_{x\to0} x\cos\left(\frac1x\right)\right)$,

then use the sqeeze theorem to evaluate $\lim\_{x\to0} x\cos(\frac1x)=0$,
            and use continuity to evaluate $\lim\_{x\to0} \frac{4}{1+x}=4$.
            
            ---
            ## Using continuity to evaluate limits
            #### Example
            Evaluate $\displaystyle \lim\_{x\to0} f(x)$, $\displaystyle \lim\_{x\to0} g(x)$, $\displaystyle \lim\_{x\to\frac\pi2} q(x)$, where
            $\displaystyle
              f(x) = \frac{4x}{x+1}\cos\left( \frac1x \right)
            $,

$\displaystyle
              g(x) = ([\operatorname{sgn}(x)]^2+5x)^2
            $,  
            $\displaystyle
              q(x) = \sin([\tan(x)]^2)
            $.

#### Solution
            By definition, $\operatorname{sgn}(0)=0$ but $\lim\_{x\to0^+}\operatorname{sgn}(x)=1$, so $\operatorname{sgn}$ is discontinuous at $0$.
            However, $[\operatorname{sgn}(x)]^2=1$ for all $x\neq0$, and the limit only sees $x$ close to $0$, not $x=0$.

Therefore, for the purpose of evaluating this limit, we can replace $[\operatorname{sgn}(x)]^2$ by the constant function $1$, which is continuous.

So $\displaystyle \lim\_{x\to0} g(x) = \lim\_{x\to0} ([\operatorname{sgn}(x)]^2+5x)^2 = \lim\_{x\to0} (1+5x)^2 = 1$, where we have used continuity of the linear function and the square function.
            
            ---
            ## Using continuity to evaluate limits
            #### Example
            Evaluate $\displaystyle \lim\_{x\to0} f(x)$, $\displaystyle \lim\_{x\to0} g(x)$, $\displaystyle \lim\_{x\to\frac\pi2} q(x)$, where
            $\displaystyle
              f(x) = \frac{4x}{x+1}\cos\left( \frac1x \right)
            $,

$\displaystyle
              g(x) = ([\operatorname{sgn}(x)]^2+5x)^2
            $,  
            $\displaystyle
              q(x) = \sin([\tan(x)]^2)
            $.

#### Solution
            Because $\tan(\frac\pi2)$ is undefined, $q$ is discontinuous at $\frac\pi2$.
            Also the left and right limits of $\tan$ at $\frac\pi2$ do not match.

However, $\displaystyle \lim\_{x\to\frac\pi2} [\tan(x)]^2=\infty$ because the left and right limits do match now.

But we still can't use continuity, because $\sin(x)$ is not defined at $x=\infty$, and we do not have a meaning for $\sin$ being continuous at $\infty$.
            
            ---
            ## Another reason to care about continuity
            #### Theorem (Intermediate value theorem)
            If $f:[a,b]\to\mathbb R$ is continuous and $y$ is strictly between $f(a)$ and $f(b)$,

then $\exists t\in(a,b)$ such that $f(t)=y$.
            
            ---
            ## Another reason to care about continuity
            #### Theorem (Intermediate value theorem)
            If $f:[a,b]\to\mathbb R$ is continuous and $y$ is strictly between $f(a)$ and $f(b)$,

then $\exists t\in(a,b)$ such that $f(t)=y$.
            #### Interpretation
            If there is a line dividing two points, and you want to draw a curve from one point to the other, then your curve will cross the line at least once.
            
            ---
            ## Another reason to care about continuity
            #### Theorem (Intermediate value theorem)
            If $f:[a,b]\to\mathbb R$ is continuous and $y$ is strictly between $f(a)$ and $f(b)$,

What happens if you evaluate $f(\pm k)$ for $k \gt 0$ large enough?
            A sketch might help.
            
            ---
            ## Another reason to care about continuity
            #### Theorem (Intermediate value theorem)
            If $f:[a,b]\to\mathbb R$ is continuous and $y$ is strictly between $f(a)$ and $f(b)$,

then $\exists t\in(a,b)$ such that $f(t)=y$.
            #### Corollary
            Every odd order polynomial has a real root.
            #### Proof
            Because $f$ is a polynomial, it is continuous.

If we make $k \gt 0$ very large, then only the highest power of $x$ in the polynomial is significant, so $f(-k)\lt0$ and $f(k)\gt0$ or the other way around.
            In either case, $0$ is strictly between $f(-k)$ and $f(k)$.

Therefore, by the intermediate value theorem, there exists some $t\in(-k,k)$ such that $f(t)=0$.