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The foundation of calculus rests on two concepts functions and limits. We have already encountered the function, which is just a one to one rule converting one set of numbers to another set of numbers. In this section, we will have our first encounter with limits, which are basically just the value a function in the neighborhood of a point. The two fundamental branches of calculus, differentiation and integration, are directly explained in terms of limits and functions. This section describing how to evaluate limits lays the additional framework for us to learn calculus.

Most of the elementary functions we will deal with in calculus are continuous. A continuous function can be

drawn without lifting your pen. Limits at places where a function is continuous are simple.

As x approaches c from the left, but does not reach c, the value of f(x) approaches f(c). This can be written more generally as a left hand limit.

This is said as, “The limit of f(x) as x approaches c from the left equals M.” In our case of a continuous function,

As x approaches the point c from the right, but does not reach c, the value of f(x) approaches f(c). This can be written more generally as a right hand limit.

This is said as, “The limit of f(x) as x approaches c from the right equals N.” In our case of a continuous function,

When the left hand limit and the right hand limit are equal N=M then we say the two sided limit exists, or simply the limit exists, and is equal to the left and right hand limits.

This is said as, “The limit of f(x) as x approaches c equals L.” When the right and left hand limits associated with a limit are not equal or are not a number (such as infinity) then we say the limit does not exist, DNE.

With a notion of the limit we can also define continuity properly. At points where a function is continuous,

If a function is continuous at a point then the limit can be evaluated there by direct substitution. Easy limit problems general boil down to determining continuity.

\begin{figure}

\includegraphics[width= 5in]{../images/Continuous.pdf}

\caption{A continuous function can be drawn without lifting your pen. At $x=c$ the function approaches $f(c)$ from both sides.}

\end{figure}

Theorem 1. The elementary functions: trigonometric, hyperbolic, polynomials, rational, exponential, logarithmic, roots, and power laws are continuous throughout their domains.

For our elementary functions, this theorem saves us a lot of time in calculating limits. We can use direct substitution for elementary functions throughout their domain. While we should be able to identify continuity by inspection, there are some other limit laws and continuity theorems that can help prove continuity and simplify limit calculations.

Theorem 2. If at the point x, f(x) and g(x) are continuous then, the following statements are true.

1. f(x) plus or minus g(x) is continuous

2. cf(x) is continuous

3. f(x)g(x) is continuous

4. f(x)/g(x) is continuous where g(x) is not zero

Theorem 3. If f(x) is continuous at

then

Example 1. Using interval notation, state where the following functions are continuous.

Solution 1. The function t(x) is a polynomial. Polynomials are continuous over the real line. The domain of continuity of x is the real line. The function u(x) is continuous for arguments 1+x>0. The domain of continuity is $latex x in (-1,\infty )&s=1$. The function $latex v(x)&s=1$ is continuous for arguments $latex 1+\sin x>0&s=1$. There are infinitely many points where $latex \sin x=-1&s=1$ and the function is not continuous there. We say the domain of continuity is $latex x\in (-\infty ,\infty )\backslash \{3\pi /2+2n\pi \} ,n\in {\mathbb Z}&s=1$. The function $latex \ln (1+\tan x)&s=1$ is continuous everywhere $latex 1+\tan x>0&s=1$. In the interval between $latex (-\pi /2,\pi /2)&s=1$ the points between $latex x\in (-\pi /2,-\pi /4]&s=1$ are removed so the domain is $latex x\in [-\pi /4,\pi /2)+n\pi ,n\in {\mathbb Z}&s=1$

Example 2. Where are the following functions continuous?

Solution 2. All of these functions are continuous everywhere. For $latex f_{1} (x)&s=1$ the polynomials are continuous over the real line. For $latex f_{2} (x)&s=1$, $latex \cos x&s=1$ is continuous over the real line, so by the composition of continuous functions theorem $latex \sin (\cos x)&s=1$ is continuous everywhere. For $latex f_{3} (x)&s=1$, the square root is everywhere non-negative. The limit as x goes to 0 from the left also exists because $latex x^{4} &s=1$ is positive and $latex 1-\sin x&s=1$ goes to one. For $latex f_{4} (x)&s=1$, $latex \arctan x&s=1$ is continuous over the real line. For $latex f_{5} (x)&s=1$ by the above theorem composition of functions are continuous. For $latex f_{5} (x)&s=1$, unlike $latex \tan x&s=1$, the hyperbolic function $latex \tanh x&s=1$ has no singularities. The functions $latex \sinh x&s=1$ and $latex \cosh x&s=1$ are continuous over the real line, so is the composition of all three functions is also. Functions like $latex f_{6} (x)&s=1$ are continuous everywhere except where the denominator equals zero. Since the denominator only equals zero at the points plus or minus $latex i&s=1$ the denominator is defined for all real numbers. For $latex f_{7}(x)&s=1$, $latex \ln (1+x^{6} )&s=1$ is continuous everywhere since the argument is positive for every x.

Continuity is more common than not. Look out for it first before starting a limit problem. Of course there are times when continuity may not be found, but these may be evil monsters. Take for example

Is this function continuous anywhere? Don’t worry we won’t be doing any calculus with this guy.