Sandwich Theorem

If $f(x),\ g(x)$ and $h(x)$ are any three functions such that,

$f(x) \leq g(x) \leq h(x)\forall x \in$ neighbourhood of $x = a$ and $\lim_{x \rightarrow a}\mspace{2mu} f(x) = \lim_{x \rightarrow a}\mspace{2mu} h(x) = I$ (say).

Then $\lim_{x \rightarrow a}\mspace{2mu} g(x) = I$ . This theorem is normally applied when the $\lim_{x \rightarrow a}\mspace{2mu} g(x)$ can not be obtained by using conventional methods as function $f(x)$ and $h(x)$ can be easily found.

IIIustration 15:

Evaluate $\lim_{n \rightarrow \infty}\mspace{2mu}\frac{\lbrack x\rbrack + \left\lbrack 2^{2} \cdot x \right\rbrack + \left\lbrack 3^{2} \cdot x \right\rbrack + \ldots\ldots\ldots\left\lbrack n^{2} \cdot x \right\rbrack}{n^{3}}$ where [.] denotes the greatest integer function denotes the greatest integer function.

Solution:

We know $x - 1 < \lbrack x\rbrack < x$

$\begin{matrix} & \ \Rightarrow x + 2^{2}x + 3^{2}x + \ldots\ldots + n^{2}x - n < \sum_{r = 1}^{n}\mspace{2mu}\mspace{2mu}\left\lbrack r^{2}x \right\rbrack < x + 2^{2}x + 3^{2}x + \ldots n^{2}x \\ & \ \Rightarrow \frac{n(n + 1)(2n + 1)}{6}x - n < \sum_{r = 1}^{n}\mspace{2mu}\mspace{2mu}\left\lbrack r^{2}x \right\rbrack < \frac{n(n + 1)(2n + 1)}{6}x \\ & \ \Rightarrow \frac{\left( 1 + \frac{1}{n} \right)\left( 2 + \frac{1}{n} \right)}{6}x - \frac{n}{n^{3}} < \frac{1}{n^{3}}\sum_{r = 1}^{n}\mspace{2mu}\mspace{2mu}\left\lbrack r^{2}x \right\rbrack < \frac{\left( 1 + \frac{1}{n} \right)\left( 2 + \frac{1}{n} \right)}{6}x \end{matrix}$

Now $= \lim_{x \rightarrow \infty}\mspace{2mu}\frac{\left( 1 + \frac{1}{n} \right)\left( \left( 2 + \frac{1}{n} \right) \right)}{6}x - \frac{1}{n^{2}} < \lim_{x \rightarrow \infty}\mspace{2mu}\frac{1}{n^{3}}\sum_{r = 1}^{n}\mspace{2mu}\left\lbrack r^{2}x \right\rbrack < \lim_{n \rightarrow \infty}\mspace{2mu}\frac{\left( 1 + \frac{1}{n} \right)\left( \left( 2 + \frac{1}{n} \right) \right)}{6}x$

Now $\lim_{n \rightarrow \infty}\mspace{2mu}\frac{\left( 1 + \frac{1}{n} \right)\left( \left( 2 + \frac{1}{n} \right) \right)}{6}x - \frac{1}{n^{2}} = \frac{x}{3},\$ and $\lim_{x \rightarrow \infty}\mspace{2mu}\frac{\left( 1 + \frac{1}{n} \right)\left( 2 + \frac{1}{n} \right)x}{6} = \frac{x}{3}$

Hence $\lim_{x \rightarrow \infty}\mspace{2mu}\frac{1}{x^{3}}\sum_{r = 1}^{n}\mspace{2mu} r^{2}x = \frac{x}{3}$

$\begin{matrix} & \text{~Alternate:~} \\ & \ \begin{matrix} \text{~we know~}\lbrack x\rbrack = x - \{ x\} \\ = \frac{n(n + 1)(2n + 1)}{6}\left\lbrack r^{2}x \right\rbrack = x\sum_{r = 1}^{n}\mspace{2mu}\mspace{2mu} r^{2} - \left( \{ x\} + \left\{ 2^{2}x \right\} + \ldots + \left\{ n^{2}x \right\} \right) \\ \Rightarrow \frac{1}{n^{3}}\sum_{r = 1}^{n}\mspace{2mu}\mspace{2mu} r^{2} = \frac{n}{6}\left( 1 + \frac{1}{n} \right)\left( 2 + \frac{1}{n} \right) - \left( \frac{\left. \ \{ x\} + \left\{ 2^{2}x \right\} + \ldots + \left\{ n^{2}x \right\} \right)}{n^{3}} \right) \\ \text{~Since~}0 \leq \left\{ r^{2}x \right\} < 1 \Rightarrow 0 \leq \sum_{r = 1}^{n}\mspace{2mu}\mspace{2mu}\left\{ r^{2}x \right\} \leq n \\ \Rightarrow \lim_{n \rightarrow \infty}\mspace{2mu}\frac{\sum_{r = 1}^{n}\mspace{2mu}\mspace{2mu}\left\{ r^{2}x \right\}}{n^{3}} = 0 \Rightarrow \lim_{n \rightarrow \infty}\mspace{2mu}\frac{1}{n^{3}}\sum_{r = 1}^{n}\mspace{2mu}\mspace{2mu}\left\{ r^{2}x \right\} = \frac{x}{3} \end{matrix} \end{matrix}$

10. Continuity of a Function

Most of the result in calculus are not true unless we are dealing with functions that are continuous. Hence its very important to understand the concept of continuity. Although the meaning of continuity seems to clear to most student, but a formal definition of continuity in term of limits is also very much needed.

A function $f(x)$ is said to be continuous at $x = a$ if $\lim_{x \rightarrow a^{-}}\mspace{2mu} f(x) = \lim_{x \rightarrow a^{+}}\mspace{2mu} f(x) = f(a)$

i.e. L.H.L. $=$ R.H.L. $=$ value of the function at a i.e. $\lim_{x \rightarrow a}\mspace{2mu} f(x) = f(a)$ .

If $f(x)$ is not continuous at $x = a$ , we say that $f(x)$ is discontinuous at $x = a$ .

Geometrical meaning: The function ‘ $f$ ‘will be continuous at $x = a$ if there is no break in the graph of the function $y = f(x)$ at the point (a, $f(a)$ ).

Reasons For Discontinuity of a Function

One of the following may be the reasons for the discontinuity of $f(x)$

(i) $\lim_{x \rightarrow a^{-}}\mspace{2mu} f(x)$ and $\lim_{x \rightarrow a^{+}}\mspace{2mu} f(x)$ exist but are not equal. For example $f(x) = \lbrack x\rbrack$ is discontinuous at all integer points.

(ii) $\lim_{x \rightarrow a^{-}}\mspace{2mu} f(x)$ and $\lim_{x \rightarrow a^{+}}\mspace{2mu} f(x)$ exist and are equal but not equal to $f(a)$ . for example $f(x) = \lbrack sinx\rbrack$ where $x \in (0,\ \pi)$ at $x = \frac{\pi}{2}$

(iii) when $f(x)$ is not defined at $x = a$ . For example $f(x) = 1/x - 1$

(iv) At least one of the limits $\lim_{x \rightarrow a^{-}}\mspace{2mu} f(x)$ or $\lim_{x \rightarrow a^{+}}\mspace{2mu} f(x)$ does not exist or atleast one of these limits is $\infty$ or $- \infty$ .

Continuity of a Function in an Interval

Continuity in an open interval:

A function $f(x)$ is said to be continuous in an open interval ( $a,\ b$ ) if it is continuous at each point of the interval ( $a,\ b$ ).

Continuity in a closed interval:

A function $f(x)$ is said to be continuous in a closed interval $\lbrack a,\ b\rbrack$ if

(i) $f(x)$ is continuous at each point of the interval $(a,\ b)$ .

(ii) $f(x)$ is continuous from right at $x = a$ i.e. $\lim_{x \rightarrow a^{+}}\mspace{2mu} f(x) = f(a)$ .

(iii) $f(x)$ is continuous from left at $x = b$ i.e. $\lim_{x \rightarrow b^{-}}\mspace{2mu} f(x) = f(b)$ and;

Geometrical meaning:

The function $f(x)$ will be continuous in the closed interval $\lbrack a,\ b\rbrack$ if the graph of $y = f(x)$ is an unbroken line (curved or straight) from the point (a, $f(a))$ to (b, $f(b)$ ).

Algebra of Continuous Functions

Let $f(x)$ and $g(x)$ be two functions, then the following results holds true.

Case I: If $f$ and g both are continuous at $x = a$ , then $c_{1}f(x) \pm c_{2}\text{ }g(x)$ and $f(x) \cdot g(x)$ will be continuous at $x = a$ . And $f(x)/g(x)$ will also be continuous at $x = a$ , provided $g(a) \neq 0$ .

Case II: When one of the function $f$ or g is discontinuous at $x = a$ , then $c_{1}f(x) \pm c_{2}\text{ }g(x)$ is definitely discontinuous, but nothing can be said about the continuity of $f(x) \cdot g(x)$ and $f(x)/g(x)$ . They may or may not be continuous at $x = a$ .

Case III: When $f$ and $\mathbf{g}$ both are discontinuous at $\mathbf{x} = \mathbf{a}$ , then nothing can be said about the continuity of $\mathbf{c}_{1}f(\mathbf{x}) \pm c_{2}\text{ }g(x),\ f(x).g(x)$ and $f(x)/g(x)$ .

Properties of a Function Continuous In A Closed Interval

If a function is continuous on a closed interval $\lbrack a,\ b\rbrack$ , then

1. It can't approach towards $\infty$

Not possible

2. If $f(\alpha):f(\beta) < 0$ , where $\alpha,\ \beta \in \lbrack a,\ b\rbrack$ , then there exist atleast $y \in \lbrack\alpha,\ \beta\rbrack$ such $f(y) = 0$

Or If $f(\alpha) \cdot f(\beta) < 0$ where $\alpha,\ \beta \in \lbrack a,\ b\rbrack$ , then we can say that atleast one root of the equation $f(x) = 0$ will lie between $(\alpha,\ \beta)$ .

3. If $\beta$ lies between $f(c)$ and $f(d)$ then will exist atleast one $p \in \lbrack c,\ d\rbrack$ such that $f(p) = \beta$ . This result is called intermediate value theorem.

4. $f(x)$ has a minimum (m) and maximum (M) value an $\lbrack a,\ b\rbrack$

Illustration 16:

Let $f:\lbrack 0,\ 1\rbrack \rightarrow \lbrack 0,\ 1\rbrack$ bijective and continuous function. Show that there exist at least one $x \in \lbrack 0,\ 1\rbrack$ such that $f(x) = x$

Solution:

Let $h(x) = f(x) - x$ . now $h(0) = f(0) \geq 0$ and $h(1) = f(0) - 1 \leq 0$

$\Rightarrow h(0) \cdot h(1) \leq 0$

Hence we can say there exist at least one $x \in \lbrack 0,\ 1\rbrack$ such that $h(x) = 0$

$\Rightarrow f(x) = x$

Continuity of Composite Functions

Let $f(x)$ and $g(x)$ are two functions, now we are interested in the continuity of $f(g(x))$ .

Case I: If $f$ and g both are continuous, then $f(g(x))$ will also be continuous.

Case II: If $f$ is continuous and g is discontinuous. Here again two cases arises.

(a) If points of discontinuity of $g(x)$ are not lying in the domain, the $f(g(x))$ will be definitely discontinuous at those points.

(b) If points of discontinuity lies in the domain, then nothing can be said about the continuity of $f(g(x))$ in general.

Case III: If $f$ and g both are discontinuous, the also nothing can be about the continuity of $f(g(x))$ .

IIIustration 17:

Find the points of discontinuity of $g(f(x))$ if $g(x) = \frac{1}{x^{2} + x - 2}$ and $f(x) = \frac{1}{x - 1}$ .

Solution:

The function $f(x) = \frac{1}{x - 1}$ is discontinuous at the point $x = 1$ .

The function $g(f(x)) = \frac{1}{f^{2}(x) + f(x) - 2} = \frac{1}{(f(x) + 2)(f(x) - 1)}$

is discontinuous at $f(x) = - 2$ and $f(x) = 1$ .

When $f(x) = - 2,\ \frac{1}{x - 1} = - 2 \Rightarrow x = \frac{1}{2}$

When $f(x) = 1,\ \frac{1}{x - 1} = 1 \Rightarrow x = 2$ .

Hence, the composite function $y = g(f(x))$ is discontinuous at three points $x = 1/2,\ 1,\ 2$ .

Limit & Continuity

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