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Saturday, 9 August 2014

chapter 8 Methods for Evaluation of Limits

Now we discuss the various methods used in obtaining limits. Each method will be accompanied by some examples illustrating that method.
(A) DIRECT SUBSTITUTION
This already finds mention at the start of the current section, where we saw that for a continuous function, the limit can be obtained by direct substitution.
This is because, by definition of a continuous function (at x = a):
{\rm{LHL}}\left( {{\rm{at}}\,x = a} \right) = {\rm{RHL}}\left( {{\rm{at}}\,x = a} \right) = f\left( a \right)
Hence, for example, all polynomial limits can be evaluated by direct substitution.
Some examples make all this clear:
(i) \mathop {\lim }\limits_{x \to 1} {x^3} + 1 = 2
(ii) \mathop {\lim }\limits_{x \to 2} 5{x^2} + 3x + 1 = 27
(iii) \mathop {\lim }\limits_{x \to  - 1} 4{x^3} + 4 = 0
(iv) \mathop {\lim }\limits_{y \to 1} |y| + 1 = 2
(v) \mathop {\lim }\limits_{x \to 5} \dfrac{{5{x^2} + 4}}{{2x + 7}} = \dfrac{{129}}{{17}}
(vi) \mathop {\lim }\limits_{x \to 8} \dfrac{{{x^3} + 1}}{{x + 1}} = \dfrac{{513}}{9} = 57
(B) FACTORIZATION
We saw an example of this method in evaluating \mathop {\lim }\limits_{x \to 1} \dfrac{{{x^2} - 1}}{{x - 1}}.

(i)\mathop {\lim }\limits_{x \to 1} \dfrac{{{x^3} - 1}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \dfrac{{(x - 1)({x^2} + x + 1)}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \left( {{x^2} + x + 1} \right) = 3
In such forms, the limit is indeterminate due to a certain factor occuring in the expression (For example, in the limit above, (x - 1) occurs in both the numerator and denominator and makes the limit indeterminate, of the form \dfrac{0}{0} ) . Factorization leads to cancellation of that common factor and reduction of the limit to a determinate form.
Note that this limit is also of the form \mathop {\lim }\limits_{x \to 1} \dfrac{{{x^m} - 1}}{{x - 1}} (whose limit is m)
(ii) \mathop {\lim }\limits_{x \to 2} \dfrac{{{x^2} - 3x + 2}}{{{x^2} - 4}} = \mathop {\lim }\limits_{x \to 2} \dfrac{{(x - 1)(x - 2)}}{{(x - 1)(x + 2)}} = \mathop {\lim }\limits_{x \to 2} \dfrac{{x - 1}}{{x + 2}} = \dfrac{1}{4}
(iii)\mathop {\lim }\limits_{x \to 0} \dfrac{{(1 + x)(1 + 2x)(1 + 3x) - 1}}{x}
 = \mathop {\lim }\limits_{x \to 0} \dfrac{{1 + 6x + 11{x^2} + 6{x^3} - 1}}{x}
\mathop {\lim }\limits_{x \to 0} \dfrac{{6x + 11{x^2} + 6{x^3}}}{x}
 = \mathop {\lim }\limits_{x \to 0} (6 + 11x + 6{x^2}) = 6

We see that upon substitution of
 x = 1, both the numerator and denominator become 0. Hence, (x - 1) is a factor of both the numerator and denominator (Factor theorem)
(iv) \mathop {\lim }\limits_{x \to 1} \,\,\dfrac{{{x^4} - 3x + 2}}{{{x^5} - 4x + 3}}
Factorization leads to
\mathop {\lim }\limits_{x \to 1} \dfrac{{{x^4} - 3x + 2}}{{{x^5} - 4x + 3}}
= \mathop {\lim }\limits_{x \to 1} \,\,\dfrac{{(x - 1){x^3} + {x^2} + x - 2}}{{(x - 1){x^4} + {x^3} + {x^2} + x - 3}}
Cancelling out (x - 1) from the numerator and the denominator.
 = \mathop {\lim }\limits_{x \to 1} \dfrac{{{x^3} + {x^2} + x - 2}}{{{x^4} + {x^3} + {x^2} + x - 3}} = \dfrac{1}{1} = 1
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