09/09/2021

Today we still focus on calculating limits and continuity.

Calculating Limits(Continued)

Let's review some of the material in the textbook first. Limits in general are very hard to compute, and what we can do is to start from well-known limits and use the rules we present here to turn any complicated functions into algebraic combinations of functions with well-known limits.

The following algebraic operations commute with limits, assuming all the limits exist:

$\lim_{x\to a} (f(x)\pm g(x))=\lim_{x\to a} f(x)\pm\lim_{x\to a} g(x)$;
$\lim_{x\to a} f(x)g(x)=\lim_{x\to a} f(x)\lim_{x\to a} g(x)$;
$\lim_{x\to a}\frac{f(x)}{g(x)} =\frac{\lim_{x\to a} f(x)}{\lim_{x\to a} g(x)}$, assuming the limit of $g$ at $a$ is non-zero.

The limit of the following functions are well-known:

$\lim_{x\to a} x^r=a^r$, assuming that $x^r$ is well-defined at $a$;
$\lim_{x\to a} \sin (x)=\sin (a)$;
$\lim_{x\to a} r^x =r^a$, for any positive real number $r$;
$\lim_{x\to a} \log_b x =\log_b a$, for any $a>0$ and $b>0$ but $b\neq 1$.
$\lim_{x\to 0}\frac{\sin x}{x} =1$.

(Squeeze Theorem)

If we have the inequality $g(x)\leq f(x)\leq h(x)$ holds in an open interval $(a-\delta ,a+\delta )$ containing $a$, where $\delta >0$ is a positive real number, and there exists $L>0$ such that $\lim_{x\to a} g(x)=\lim_{x\to a} h(x)=L$, then we have $\lim_{x\to a} f(x)=L$.

You can imagine this intuitively: if a real number $x$ satisfies $a\leq x\leq a$, then $x$ must be $a$. Here we just present a limit version of this tautological argument.

Now we explore some examples.

Compute the limit $\displaystyle\lim_{x\to 4}\frac{x^2 -4x}{x^2-3x-4}$.

We can firstly try to substitute $x=4$ to both the numerator and the denominator. Write $\displaystyle f(x)=\frac{x^2 -4x}{x^2 -3x-4}=\frac{g(x)}{h(x)}$, then we have $g(4)=0$ and $h(4)=0$, so we cannot directly substitute $x$ by $4$ to get the limit and it requires further observation. Note that $g(x)=x^2 -4x=x(x-4)$ and $h(x)=x^2 -3x-4 =(x-4)(x+1)$, so we have $$\lim_{x\to 4} f(x)=\lim_{x\to 4}\frac{g(x)}{h(x)} =\lim_{x\to 4}\frac{x(x-4)}{(x+1)(x-4)} =\lim_{x\to 4}\frac{x}{x+1} =\frac{4}{5}.$$

Continuity

Continuity is a concept related to limits. We say a function $f(x)$ is continuous at a point $a$ if it's defined at $a$ and $\displaystyle\lim_{x\to a} f(x)=f(a)$. There're two useful Corollaries from continuity, one can help us compute some more limits, and one can tell us the existence of roots.

If $f$ is continuous and $g$ has limit at $a$ for some real number $a$, then$$\lim_{x\to a} f\circ g(x)=f(\lim_{x\to a} g(x)).$$

(Intermediate Value Theorem)

Let $f$ be continuous in the interval $[a,b]$. If $f(a)f(b)< 0$, then $f$ must have a root in $(a,b)$.

Find the limit $\lim_{x\to 0}\cos (x+\sin x)$.

We can decompose the function as $f(g(x))$, where $f(y)=\cos y$ and $g(x)=x+\sin x$. Note that $g(x)$ and $f(x)$ are all continuous, we can apply Theorem 5.4 to this function and obtain that $$\lim_{x\to 0}\cos (x+\sin x)=\cos (\lim_{x\to 0} (x+\sin x))=\cos (0)=1.$$

Show that there's a root of the following equation:

$2\sin x=3-2x$;$x^5-x^3+3x-5=0$.

For the first one, let $f(x)=2\sin x-3+2x$. If we plug in values $0$ and $1$, then we get $f(0)=-3$ and $f(1)=2\sin 1-1$. Note that $\sin 1>\sin\frac{\pi}{6} =\frac{1}{2}$, so $f(1)>0$, and hence by the Intermediate Value Theorem, $f$ has a root in the interval $(0,1)$. This means that there is a real number $0 < a< 1$ so that $f(a)=0$, which implies that $2\sin a=3-2a$, i.e. $a$ solves the equation $2\sin x=3-2x$.

For the second one, let $f(x)=x^5-x^3+3x-5$, then we have $f(0)=-5$ and $f(2)=25$, hence by the Intermediate Value Theorem, $f$ should have a root in the interval $(0,2)$.