Let us consider a simple geometry problem. Given a curve $y=f(x)$, we want to find a line tangent to the graph of $y=f(x)$ at $x=a$ (meaning the line meets the graph of $y=f(x)$ exactly at a point $(a,f(a))$ on a small interval containing $x=a$.

One may wonder at this point why finding a tangent line is a big deal. Well, it is in fact a pretty big deal besides mathematicians’ purely intellectual curiosities. There is a reason why Sir Issac Newton had to invent calculus of which crucial notion is the slope of a tangent line. It is still too early to talk about why it is important or useful. We will get there when we are ready.

We attempt to tackle the problem with an example first. Here is an example we want to consider

Example. Find the equation of a line tangent to the graph of $y=x^2$ at $x=2$.

Solution. To find the equation of a line, we need two ingredients: slope and $y$-intercept or slope and a point. We already know a point. We know that the line must pass through $(2,4)$. So all we need to find is its slope $m$. From algebra, we know that the equation of a line passing through $(2,4)$ with slope $m$ is given by $y-4=m(x-2)$ or $y=mx-2m+4$. Since $y=x^2$ and $y=mx-2m+4$ meet exactly at one point, the quadratic equation $x^2=mx-2m+4$ or $x^2-mx+2m-4=0$ must have exactly one solution. We have learned from the theory of quadratic equations that in that case the discriminant $D=b^2-4ac$ must be equal to $0$. That is, in our case $$D=m^2-4(2m-4)=m^2-8m+16=(m-4)^2=0.$$ Hence we determine that $m=4$ and the equation of the tangent line is $y=4x-4$.

So we see that finding the slope of a tangent line is not that difficult and that it does not require any new mathematics, or does it? Remember that we have not yet tackled our problem in general context. Before we get more ambitious, consider another example with a more complicated function, say $y=x^5$. Let us say that we want to find the line tangent to the graph of $y=x^5$ at $x=1$. Then the equation of the tangent line would be $y=mx-m+1$. In order for $y=x^5$ and the line $y=mx-m+1$ to meet exactly at one point, the quintic equation $x^5-mx+m-1=0$ must have exactly one solution. Our problem here is that we have no algebraic means, such as quadratic formula or discriminant, to use to determine the value of $m$. We are stuck here and there is no hope of tackling our simple geometry problem using only algebra. That is the reason why we have to cleverly devise a new way to tackle the problem. This is where we enter the realm of Calculus. The new idea to tackle the problem is not really new and it was already used by the ancient Greeks. And the world had to wait until it was rediscovered independently by Sir Issac Newton and by Gottfried Leibniz. I do not know if any of them actually knew about the ancient Greek idea.

So far we have mainly studied finite limits. Here we would like to discuss infinite limits. You may wonder why  we need to study infinite limits. They in fact do have important applications. One immediate application is that it provides  us some information on the shape of a function, i.e. it helps us sketch the graph of a function, as we will see later.

We first begin with the notion of vertical asymptotes.

Definition. The line $x=a$ is called a  vertical asymptote of the graph of $y=f(x)$ if $$\lim_{x\to a+}f(x)=\pm\infty,\ \mbox{or}\ \lim_{x\to a-}f(x)=\pm\infty.$$

Example. Find the vertical asymptotes of the graph of $y=\displaystyle\frac{x^2-3x+2}{x^3-4x}$.

Solution. The candidates for vertical asymptotes are the values of $x$ that make the denominator $0$. In our example, they are the root of the equation $x^3-4x=0$. Since $x^3-4x=x(x^2-4)=x(x+2)(x-2)$, we find three roots $x=-2,0,2$. However, some of them may not necessarily be vertical asymptotes. To check this, we calculate the limits: \begin{eqnarray*}\lim_{x\to 0+}\frac{x^2-3x+2}{x^3-4x}&=&\frac{2}{0-}\ \mbox{(Can you see why?)}\\&=&-\infty,\\\lim_{x\to 0-}\frac{x^2-3x+2}{x^3-4x}&=&\frac{2}{0+}\\&=&\infty,\\\lim_{x\to -2-}\frac{x^2-3x+2}{x^3-4x}&=&\frac{12}{0-}\\&=&-\infty,\\\lim_{x\to -2+}\frac{x^2-3x+2}{x^3-4x}&=&\frac{12}{0+}\\&=&\infty,\\\lim_{x\to 2}\frac{x^2-3x+2}{x^3-4x}&=&\lim_{x\to 2}\frac{(x-1)(x-2)}{x(x+2)(x-2)}\\&=&\lim_{x\to 2}\frac{x-1}{x(x+2)}\\&=&\frac{1}{8}.\end{eqnarray*}

So, we see that $x=0,-2$ are vertical asymptotes while $x=2$ is not.

Definition. A line $y=b$ is called a horizontal asymptote of the graph of $y=f(x)$ if $$\lim_{x\to\infty}f(x)=b,\ \mbox{or}\ \lim_{x\to -\infty}f(x)=b.$$

Example. Find the horizontal asymptotes of the graph of $y=\displaystyle\frac{5x^2+8x-3}{3x^2+2}$.

Solution. You can notice at once that the limit $\displaystyle\lim_{x\to\infty}\frac{5x^2+8x-3}{3x^2+2}$ is an $\frac{\infty}{\infty}$ type indeterminate form. So how do we calculate this kind of indeterminate form? First divide the numerator and the denominator by the highest power of $x$ appeared in the denominator:\begin{eqnarray*}\lim_{x\to\infty}\frac{5x^2+8x-3}{3x^2+2}&=&\lim_{x\to\infty}\frac{\frac{5x^2+8x-3}{x^2}}{\frac{3x^2+2}{x^2}}\\&=&\lim_{x\to\infty}\frac{5+\frac{8}{x}-\frac{3}{x^2}}{3+\frac{2}{x^2}}\\&=&\frac{5}{3}.\end{eqnarray*} The final answer is obtained by the limits $\displaystyle\lim_{x\to\infty}\frac{1}{x^n}=0,$ where $n$ is a positive integer.

Similarly, $\displaystyle\lim_{x\to\infty}\frac{5x^2+8x-3}{3x^2+2}=\frac{5}{3}$ using the limit $\displaystyle\lim_{x\to -\infty}\frac{1}{x^n}=0,$ where $n$ is a positive integer. The following picture contains the graphs of the function (in blue) and the horizontal asymptote (in red).

Figure 1. The graph of y=(5x^2+8x-3)/(3x^2+2)

Example. Find the horizontal asymptotes of $y=\displaystyle\frac{x^2-3x+2}{x^3-4x}$.

Solution. The limits $\displaystyle\lim_{x\to\pm\infty}\frac{x^2-3x+2}{x^3-4x}$ is $\frac{\infty}{\infty}$ type indeterminate form. So as we did in the previous example, we first divide the numerator and the denominator by the highest power of $x$ that is appeared in the denominator: \begin{eqnarray*}\lim_{x\to\infty}\frac{x^2-3x+2}{x^3-4x}&=&\lim_{x\to\infty}\frac{\frac{x^2-3x+2}{x^3}}{\frac{x^3-4x}{x^3}}\\&=&\lim_{x\to\infty}\frac{\frac{1}{x}-\frac{3}{x^2}+\frac{2}{x^3}}{1-\frac{4}{x^2}}\\&=&0.\end{eqnarray*}

Similarly you find that $\displaystyle\lim_{x\to -\infty}\frac{x^2-3x+2}{x^3-4x}=0$.

The following picture shows you the graph of the function (in blue), the horizontal and the vertical asymptotes (in red).

Figure 2. The graph of y=(x^2-3x+2)(x^3-4x)

Normally the graph of a function $y=f(x)$ never touches or crosses its horizontal asymptote while it gets closer and closer to its horizontal asypmtote as $x\to\infty$ or $x\to -\infty$. But there are exceptions as shown in the following example.

Example. Consider the function $f(x)=2+\displaystyle\frac{\sin x}{x}$. Using the Sandwich Theorem, one can show that $\displaystyle\lim_{x\to\pm\infty}\frac{\sin x}{x}=0$ and hence $\displaystyle\lim_{x\to\pm\infty}f(x)=2$. That is $y=2$ is a horizontal asymptote of the curve on both left and right. As you can see in the following picture, the graph crosses the horizontal asymptote $y=2$ infinitely many
times.

Figure 3. The graph of y=2+sin(x)/x.

There is another kind of asymptotes, called oblique (slanted) asymptotes. An oblique asymptote can be seen from a rational function $\frac{p(x)}{q(x)}$ where $\deg p(x)>\deg q(x)$. An oblique asymptote is in fact given by a dominating term of a rational function as you can see in the following example.

Example. Consider the rational function $f(x)=\displaystyle\frac{2x^2-3}{7x+4}$. By long division, we obtain \begin{eqnarray*}f(x)&=&\frac{2x^2-3}{7x+4}\\&=&\left(\frac{2}{7}x-\frac{8}{49}\right)+\frac{-115}{49(7x+4)}.\end{eqnarray*} As $x\to\pm\infty$, the remainder $\displaystyle\frac{-115}{49(7x+4)}\to 0$. Hence the graph of $f(x)$ gets closer to the graph of the linear function $y=\displaystyle \frac{2}{7}x-\frac{8}{49}$ as $x\to\pm\infty$. This linear function is an oblique (slanted) asymptote of the graph of $f(x)$. The following picture shows the graph of $f(x)$ (in blue) and both the vertical asymptote $x=-\frac{4}{7}$ and the oblique asymptote $y=\frac{2}{7}x-\frac{8}{49}$ (in red).

Figure 4. The graph of y=(2x^3-3)/(7x+4).

A closer look.

Figure 5. The graph of y=(2x^3-3)/(7x+4).

Intuitively speaking, we say a function is continuous at a point if its graph has no separation, i.e. there is no hole or breakage, at that point. Such notion of continuity can be defined explicitly as follows.

Definition: A function \(f(x)\) is said to be continuous at a point \(x=a\) if \[\lim_{x\to a}f(x)=f(a).\]

Note that the above definition assumes the existence of both \(\displaystyle\lim_{x\to a}f(x)\) and \(f(a)\).

There are 3 different types of discontinuities.

• \(f(a)\) is not defined.

For example, consider the function\[f(x)=\frac{x^2-4}{x-2}.\] Clearly \(f(2)\) is not defined. However the limit \(\displaystyle\lim_{x\to 2}f(x)\) exists:\begin{eqnarray*}\lim_{x\to 2}\frac{x^2-4}{x-2}&=&\lim_{x\to 2}\frac{(x+2)(x-2)}{x-2}\\&=&\lim_{x\to 2}(x+2)=4.\end{eqnarray*} As a result the graph has a hole.

This kind of discontinuity is called a removable discontinuity, meaning that we can extend \(f(x)\) to a function which is continuous at \(x=a\) in the following sense: Define \(g(x)\) by\[g(x)=\left\{\begin{array}{ccc}f(x)\ \mbox{if}\ x\ne a,\\\lim_{x\to a}f(x)\ \mbox{if}\ x=a.\end{array}\right.\]Then \(g(x)\) is a continuous at \(x=a\). The function \(g(x)\) is called the continuous extension of \(f(x)\). What we just did is basically filling the hole and the filling is the limit \(\displaystyle\lim_{x\to a}f(x)\). For the above example, we define\[g(x)=\left\{\begin{array}{ccc}\frac{x^2-4}{x-2} &\mbox{if}& x\ne 2,\\4 &\mbox{if}& x=2.\end{array}\right.\] Then \(g(x)\) is continuous at \(x=2\) and in fact, it is identical to \(x+2\).

• \(\displaystyle\lim_{x\to a}f(x)\) deos not exist.

Example. Let \(f(x)=\left\{\begin{array}{cc}2x-2,\ &1\leq x<2\\3,\ &2\leq x\leq 4.\end{array}\right.\) \(f(2)=3\) but \(\displaystyle\lim_{x\to 2}f(x)\) does not exist because \(\displaystyle\lim_{x\to 2-}f(x)=2\) while \(\displaystyle\lim_{x\to 2+}f(x)=3\).

• \(f(a)\) is defined and \(\displaystyle\lim_{x\to a}f(x)\) exists, but \(\displaystyle\lim_{x\to a}f(x)\ne f(a)\).

Example. Let \(f(x)=\left\{\begin{array}{cc}\displaystyle\frac{x^2-4}{x-2},\ &x\ne 2\\3,\ &x=2.\end{array}\right.\) Then \(f(2)=3\) and \(\displaystyle\lim_{x\to 2}f(x)=4\).

From the properties of limits (Theorem 1, Lecture 4), we obtain the following properties of continuous functions.

Theorem 9. If functions \(f(x)\) and \(g(x)\) are continuous at \(x=a\), then

1. \((f\pm g)(x)=f(x)\pm g(x)\) is continuous at \(x=a\).
2. \(f\cdot g(x)=f(x)\cdot g(x)\) is continuous at \(x=a\).
3. \(\displaystyle\frac{f}{g}(x)=\frac{f(x)}{g(x)}\) is continous at \(x=a\) provided \(g(a)\ne 0\).

There are some important classes of continous functions.

• Every polynomial function \(p(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_0\) is continuous everywhere, because \(\displaystyle\lim_{x\to a}p(x)=p(a)\) for any \(-\infty<a<\infty\).
• If \(p(x)\) and \(q(x)\) are polynomials, then the rational function \(\displaystyle\frac{p(x)}{q(x)}\) is continuous wherever it is defined \(\{x\in\mathbb{R}|q(x)\ne 0\}\).
• \(y=\sin x\) and \(y=\cos x\) are continuous everywhere.
• \(y=\tan x\) is continous where it is defined, i.e. everywhere except at the points \(x=\pm\frac{\pi}{2},\pm\frac{3\pi}{2},\pm\frac{5\pi}{2},\cdots\).
• If \(n\) is a positive integer, then \(y=\root n\of{x}\) is continuous where it is defined. That is, if \(n\) is an odd integer, it is defined everywhere. If \(n\) is an even integer,it is defined on \([0,\infty)\), the set of all non-negative real numbers.

Recall that the composite function \(g\circ f(x)\) of two functions \(f(x)\) and \(g(x)\) (read \(f\) followed by \(g\)) is defined by \[g\circ f(x):=g(f(x)).\]

Theorem 10. Suppose that \(\displaystyle\lim_{x\to a}f(x)=L\) exists and \(g(x)\)is continuous function at \(x=L\). Then\[\lim_{x\to a}g\circ f(x)=g(\lim_{x\to a}f(x)).\]

It follows from Theorem 10 that the composite function of two continuous functions is again a continuous function.

Corollary 11. If \(f(x)\) is continuous at \(x=a\) and \(g(x)\) is continuous at \(f(a)\), the the composite function \(g\circ f(x)\) is continuous at \(x=a\).

Example. The function\(y=\sqrt{x^2-2x-5}\) is the composite function  \(g\circ f(x)\) of two functions \(f(x)=x^2-2x-5\) and \(g(x)=\sqrt{x}\). The function \(f(x)=x^2-2x-5\) is continuous everywhere while \(g(x)=\sqrt{x}\) is continuous on \([0,\infty)\), so by Corollary 11, the composite function \(g\circ f(x)=\sqrt{x^2-2x-5}\) is continuous on its repective domain which is \((-\infty,1-\sqrt{6}]\) or \([1+\sqrt{6},\infty)\). The following picture shows you the graph of \(y=\sqrt{x^2-2x-5}\) on the intervals  \((-\infty,1-\sqrt{6}]\) and \([1+\sqrt{6},\infty)\).

Continuous functions exhibit many nice properties. I would like to introduced a couple of them here. The first is the so-called Max-Min Theorem.

Theorem 12. [Max-Min Theorem] If \(f(x)\) is a continuous function on a closed interval \([a,b]\), \(f(x)\) attains its maximum value and minimum value on \([a,b]\).

Another important property is the so-called Intermediate Value Theorem (IVT). The IVT has an important application in the study of equations.

Theorem 13. [The Intermediate Value Theorem] If \(f(x)\) is continuous on a closed interval \([a,b]\) and \(f(a)\ne f(b)\), then , then \(f(x)\) takes on every value between \(f(a)\) and \(f(b)\). In other words, if \(f(a)<k<f(b)\) (assuming that \(f(a)<f(b)\)), then \(f(c)=k\) for some number \(a<c<b\).

It follows from Theorem 13 that

Corollary 14. If \(f(x)\) is continuous on a closed interval \([a,b]\) and \(f(a)\cdot f(b)<0\), then \(f(x)=0\) for some \(a<x<b\).

Using  Corollary 14 we can tell if a root of the equation \(f(x)=0\) can be found in some interval. For instance

Example. Show that the equation \(x^3-x-1=0\) has a root in the interval \([-1,2]\).

Solution. Let \(f(x)=x^3-x-1\). Then \(f(x)\) is continuous on \([-1,2]\). Since \(f(-1)=-1\) and \(f(2)=5\) have different signs, by Corollary 14 there is a root of \(x^3-x-1=0\) in the open interval \((-1,2)\).

In this posting, we discuss limits of trigonometric functions. The most basic trigonometric functions are of course \(y=\sin x\) and \(y=\cos x\). They have the following limit properties.

Theorem 5. For any \(a\in\mathbb R\), \[\lim_{x\to a}\sin x=\sin a,\ \lim_{x\to a}\cos x=\cos a.\]

You notice that both \(y=\sin x\) and \(y=\cos x\) satisfy the same limit property as polynomial functions (Theorem 2 in Lecture 4). This is not a coincidence and the reason behind this is that polynomial functions, \(y=\sin x\) and \(y=\cos x\) are continuous functions. This will become clear when we discuss the continuity of a function later. Limit properties of other trigonometric function will stem out automatically from the above Theorem 5 and Theorem 1 in Lecture 4. For example, the limit property of \(y=\tan x\) is given by \[\lim_{x\to a}\tan x=\lim_{x\to a}\frac{\sin x}{\cos x}=\frac{\sin a}{\cos a}=\tan a,\] where \(\tan a\) is defined or equivalently \(\cos a\ne 0\).

Theorem 6. Suppose that \(f(x)\leq g(x)\) near \(x=a\) and both \(\displaystyle\lim_{x\to a}f(x)\), \(\displaystyle\lim_{x\to a}g(x)\) exist. Then \[\lim_{x\to a}f(x)\leq \lim_{x\to a}g(x).\]

Corollary 7. [Squeeze Theorem, Sandwich Theorem] Suppose that \(f(x)\leq g(x)\leq h(x)\) near \(x=a\). If \(\displaystyle\lim_{x\to a}f(x)=\lim_{x\to a}h(x)=L\) then \[\lim_{x\to a}g(x)=L.\]

Squeeze Theorem is useful to calculate certain type of limits such as the following example.

Example. Find the limit \(\displaystyle\lim_{x\to 0}x^2\sin\frac{1}{x}\).

Solution. Since \(-1\leq\sin\frac{1}{x}\leq 1\), \[-x^2\leq x^2\sin\frac{1}{x}\leq x^2\] for all \(x\ne 0\). Since \(\displaystyle\lim_{x\to 0}(-x^2)=\lim_{x\to 0}x^2=0\), by Squeeze Theorem \[\lim_{x\to 0}x^2\sin\frac{1}{x}=0.\] The following picture also confirms our result.

There is another important limit that involves a trigonometric function.  It is

Theorem 8. \(\displaystyle\lim_{x\to 0}\frac{\sin x}{x}=1\).

This is an important formula. You will readily see that this limit is \(\frac{0}{0}\) type indeterminate form. So this means that \(\sin x\) must have a factor  \(x\) in it. But how do we factor \(\sin x\)? It is not a polynimial! In fact. it is (sort of). This is something you are going to learn in Calculus 3 (MAT 169) but I want you to taste it. The function \(\sin x\) is can be written as a never-ending polynomial (such a polynomial is called a power series in mathematics) \[\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots,\]where \(n!\) denotes the \(n\) factorial\[n!=n(n-1)(n-2)(n-3)\cdots3\cdot 2\cdot 1.\] So \begin{eqnarray*}\lim_{x\to 0}\frac{\sin x}{x}&=&\lim_{x\to 0}\frac{x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots}{x}\\&=&\lim_{x\to 0}\left(1-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+\cdots\right)\\&=&1.\end{eqnarray*}

We have now confirmed that the formula is indeed correct, but is there a more fundamental proof without using power series? Yes, there is. In fact it can be proved using trigonometry. First consider the case when \(x\to 0+\). In this case, without loss of generality we may assume that \(x\) is an acute angle so we have the following picture.

The areas of \(\triangle OAC\), \(\sphericalangle OBC\) and \(\triangle OBD\) are, respectively, given by \(\frac{1}{2}\cos x\sin x\), \(\frac{1}{2}x\) and \(\frac{1}{2}\tan x\). Clearly from the picture they satisfy the inequality \[\frac{1}{2}\cos x\sin x<\frac{1}{2}x<\frac{1}{2}\tan x.\] Dividing this inequality by \(\frac{1}{2}\sin x\) (note that \(\sin x>0\) since \(x\) is an acute angle) we obtain\[\cos x<\frac{x}{\sin x}<\frac{1}{\cos x}\] or equivalently,\[\frac{1}{\cos x}<\frac{\sin x}{x}<\cos x.\] Now \(\displaystyle\lim_{x\to 0+}\cos x=\lim_{x\to 0+}\frac{1}{\cos x}=1\), so by Squeeze Theorem,\[\lim_{x\to 0+}\frac{\sin x}{x}=1.\] Similarly, we can also show that\[\lim_{x\to 0-}\frac{\sin x}{x}=1.\] Hence completes the proof.

Example. Find $\displaystyle\lim_{x\to 0}\frac{\sin 7x}{4x}$.

Solution. \begin{eqnarray*}\lim_{x\to 0}\frac{\sin 7x}{4x}&=&\lim_{x\to 0}\frac{7}{4}\frac{\sin 7x}{7x}\\&=&\frac{7}{4}\lim_{x\to 0}\frac{\sin 7x}{7x}\\&=&\frac{7}{4}\ \left(\lim_{x\to 0}\frac{\sin 7x}{7x}=1\right).\end{eqnarray*}

Example. Find $\displaystyle\lim_{\theta\to 0}\frac{\cos\theta-1}{\theta}$.

Solution. \begin{eqnarray*}\lim_{\theta\to 0}\frac{\cos\theta-1}{\theta}&=&\lim_{\theta\to 0}\frac{\cos\theta-1}{\theta}\frac{\cos\theta+1}{\cos\theta+1}\\&=&\lim_{\theta\to 0}\frac{\cos^2\theta-1}{\theta(\cos\theta+1)}\\&=&\lim_{\theta\to 0}\frac{\cos^2\theta-1}{\theta(\cos\theta+1)}\\&=&\lim_{\theta\to 0}\frac{-\sin^2\theta}{\theta(\cos\theta+1)}\\&=&-\lim_{\theta\to 0}\frac{\sin\theta}{\theta}\frac{\sin\theta}{\cos\theta+1}\\&=&-\lim_{\theta\to 0}\frac{\sin\theta}{\theta}\cdot\lim_{\theta\to 0}\frac{\sin\theta}{\cos\theta+1}\\&=&-1\cdot 0=0.\end{eqnarray*}