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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.image showing that approaching can be done in two ways either from the left hand side or from the right hand side
graph of an inverse function
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.
graph of greatest integer function
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
graph of greatest integer function
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 2x 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}
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