Saturday, 9 August 2014

chapter 2 Introduction to Limits

Hence, we see that a limit describes the behaviour of some quantity that depends on an independent variable, as that independent variable ‘approaches’ or ‘comes close to’ a particular value.
For example, how does $\dfrac{1}{x}$ behave when $x$ becomes larger and larger? $\dfrac{1}{x}$becomes smaller and smaller and ‘tends’ to $0$.
We write this as
 $\mathop {\lim }\limits_{x \to \infty } \dfrac{1}{x} = 0$
How does $\dfrac{1}{x}$ behave when $x$ becomes smaller and smaller and approaches $0?$$\dfrac{1}{x}$ obviously becomes larger and larger and ‘tends’ to infinity.
We write this as:
 $\mathop {\lim }\limits_{x \to 0} \dfrac{1}{x} = \infty$
The picture is not yet complete. In the example above, $x$ can ‘approach’ $0$ in two ways, either from the left hand side or from the right hand side:
 $x \to {0^ - }$: approach is from left side of $0$ $x \to {0^ + }$: approach is from right side of $0$
How do we differentiate between the two possible approaches? Consider the graph of $f\left( x \right) = \dfrac{1}{x}$ carefully.
As we can see in the graph above, as $x$ increase in value or as $x \to \infty ,f\left( x \right)$decreases in value and approaches $0$ (but it remains positive, or in other words, it approaches $0$ from the positive side)
This can be written
 $\mathop {\lim }\limits_{x \to \infty } f\left( x \right) = {0^ + }$
Similarly,
 $\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = + \infty$
What if $x$ approaches $0$, but from the left hand side $\left( {x \to {0^ - }} \right)?$ From the graph, we see that as $x \to {0^ - },\dfrac{1}{x}$ increases in magnitude but it also has a negative sign, that is $\dfrac{1}{x} \to - \infty .$
What if $x \to - \infty ?\dfrac{1}{x}$ decreases in magnitude (approaches $0$) but it still remains negative, that is, $\dfrac{1}{x}$ approaches $0$ from the negative side or $\dfrac{1}{x} \to {0^ - }$
These concepts and results are summarized below:
 (i) $\mathop {\lim }\limits_{x \to \infty } \dfrac{1}{x} = {0^ + }$ $\dfrac{1}{x}$ approaches $0$ from the positive side (ii) $\mathop {\lim }\limits_{x \to \infty } \dfrac{1}{x} = {0^ - }$ $\dfrac{1}{x}$ approaches $0$ from the negative side (iii) $\mathop {\lim }\limits_{x \to {0^ - }} \dfrac{1}{x} = - \infty$ $\dfrac{1}{x}$ remains negative and increase in magnitude (iv) $\mathop {\lim }\limits_{x \to {0^ + }} \dfrac{1}{x} = + \infty$ $\dfrac{1}{x}$ remains positive and increase in magnitude
Lets consider another example now. We analyse the behaviour of $f(x) = [x]$, as $x$ approaches $0$.
What happens when $x$ approaches $0$ from the right hand side? We see that $[x]$remains $0$.
What happens when $x$ approaches $0$ from the left hand side? $[x]$ has a value $- 1$.
Note that we are not talking about what value $[x]$ takes at $x = 0$. We are concerned with the behaviour of $[x]$ in the neighbourhood of $x = 0$, that is, to the immediate left and right of $x = 0$.
Hence, we have:
 $\mathop {\lim }\limits_{x \to {0^ + }} \left[ x \right] = 0$ $\mathop {\lim }\limits_{x \to {0^ - }} \left[ x \right] = - 1$ $\left. \begin{array}{l} \mathop {\lim }\limits_{x \to {I^ + }} \left[ x \right] = I\\ \mathop {\lim }\limits_{x \to {I^ - }} \left[ x \right] = I - 1 \end{array} \right\}I\,{\rm{is}}\,{\rm{any}}\,{\rm{integer}}$
What about $f(x) = \{ x\} ?$ This should be straightforward now:
We have now seen three different quantities regarding $f(x)$
 $\mathop {\lim }\limits_{x \to {0^ + }} \left\{ x \right\} = 0$ $\mathop {\lim }\limits_{x \to {0^ - }} \left\{ x \right\} = 1$ $\left. \begin{array}{l} \mathop {\lim }\limits_{x \to {I^ + }} \left\{ x \right\} = 0\\ \mathop {\lim }\limits_{x \to {I^ - }} \left\{ x \right\} = 1 \end{array} \right\}\,I{\rm{is}}\,{\rm{any}}\,{\rm{integer}}\,$
 (i) The left hand limit $(LHL)$at $x = a:\mathop {\lim }\limits_{x \to {a^ - }} f(x)$ Describe the behaviour of $f(x)$to the immediate left of $x=a$ (ii) The right hand limit $(RHL)$at $x = a:\mathop {\lim }\limits_{x \to {a^ + }} f(x)$ Describe the behaviour of $f(x)$to the immediate left of $x=a$ (iii) The value of $f(x)$ at$x=a:f(a)$ Give the precise value that $f(x)$takes at $x=a$
Its possible that the value of the function at $x = a$ is undefined, and yet the $LHL$ or $RHL$ (or both) exist.
For example, consider
 $f\left( x \right) = \dfrac{{{x^2} - 1}}{{x - 1}}$
$f(x)$ is clearly not defined at $x = 1$.
Every where else, $f(x)$ can be written in a simple form as
 $f\left( x \right) = \dfrac{{\left( {x - 1} \right)\left( {x + 1} \right)}}{{\left( {x - 1} \right)}} = x + 1$
which is a straight line
This line has a hole at $x = 1$ because $f(x)$ is undefined there. In the neighbourhood of
 $x = 1\left( {\,{\rm{as}}\,x \to {1^ + }\,{\rm{or}}\,x \to {1^ - }} \right),$
$f(x)$ ‘approaches’ the value $2$ (though it never becomes $2$, because to become $2$$x$ has to have the value $1$, which is not possible). We see that
 ${\rm{LHL}} = \mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = 2 = \mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = {\rm{RHL}}$
To emphasize once again, in evaluating a limit at $x = a$, we are not concerned with what value $f(x)$ assumes at precisely $x = a$; we are concerned with only how $f(x)$ behaves as $x$ approaches or nearly becomes $a$, whether from the left hand or right hand side, giving rise to $LHL$ and $RHL$ respectively.
And finally, the limit of $f(x)$ at $x = a$ is said to exist if the function approaches the same value from both sides
 $\begin{array}{l} {\rm{LHL}}\,{\rm{ = }}\,{\rm{RHL}} \,\,\,\,\,\, {\rm{at}}\,\,x = a\\ {\rm{implies}}\,\,\,\,\, \mathop {\lim }\limits_{ x \to a} f\left( x \right)\,\,{\rm{exists}}\\ {\rm{and}} \,\,\,\,\,\mathop {\lim }\limits_{x \to a} f\left( x \right) = {\rm{LHL = RHL}} \end{array}$