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Saturday, 9 August 2014

chapter 10 Worked Out Examples

     Example: 1  

Evaluate the following limits
(a) \mathop {\lim }\limits_{x \to 2} \left( {\dfrac{2}{{x(x - 2)}} - \dfrac{1}{{{x^2} - 3x + 2}}} \right)
(b) \mathop {\lim }\limits_{n \to \infty } \left[ {(1 + x)(1 + {x^2})(1 + {x^4})\ldots(1 + {x^{{2^n}}})} \right]|x|< 1
(c) \mathop {\lim }\limits_{x \to 3} \dfrac{{{x^3} - 7{x^2} + 15x - 9}}{{{x^4} - 5{x^2} + 27x - 27}}
Solution: 1-(a)

This limit is of the indeterminate form \infty  - \infty . Combining the two dfractions in this limit should lead to a cancellation of the factor giving rise to this indeterminacy, i.e. (x - 2)
\mathop {\lim }\limits_{x \to 2} \left( {\dfrac{2}{{x(x - 2)}} - \dfrac{1}{{{x^2} - 3x + 2}}} \right)
\mathop {\lim }\limits_{x \to 2} \dfrac{{2(x - 1) - x}}{{x(x - 2)(x - 1)}} = \mathop {\lim }\limits_{x \to 2} \dfrac{{x - 2}}{{x(x - 1)(x - 2)}}
Cancelling out (x - 2) from the numerator and the denominator.
 = \mathop {\lim }\limits_{x \to 2} \dfrac{1}{{x(x - 1)}} = \dfrac{1}{2}
Solution: 1-(b)

Before trying to solve this, try to feel that this expression will have a finite limit even though the number of factors being multiplied tends to infinity. This is because the successive factors become closer and closer to 1 and their ‘contribution’ to the final product becomes smaller and smaller Now, to simplify this product, we multiply it by \dfrac{{1 - x}}{{1 - x}}. This is what happens:
\mathop {\lim }\limits_{n \to \infty } \left[ {\dfrac{{\overbrace{(1 - x).(1 + x)}(1 + {x^2})(1 + {x^4})\ldots(1 + {x^{{2^n}}})}}{{(1 - x).}}} \right]
\mathop {\lim }\limits_{n \to \infty } \left[ {\dfrac{{\overbrace{(1 - {x^2})(1 + {x^2})}(1 + {x^4})\ldots (1 + {x^{{2^n}}})}}{{(1 - x).}}} \right]
\mathop {\lim }\limits_{n \to 0} \left[ {\dfrac{{\overbrace{(1 - {x^4})(1 + {x^4})}\ldots (1 + {x^{{2^n}}})}}{{(1 - x).}}} \right]
\vdots
= \mathop {\lim }\limits_{n \to \infty } \,\,\dfrac{{1 - {x^2}^{n + 1}}}{{1 + x}}
Since, |x| < 1,\mathop {\lim }\limits_{n \to \infty } {x^{{2^{n + 1}}}} = 0

Hence, the value of the limit is
\dfrac{1}{{1 - x}}
Solution: 1-(c)

The numerator and denominator both tend to 0 as x \to 3 because of the common factor (x - 3).
Hence, factorization leads to :
\mathop {\lim }\limits_{x \to 3} \dfrac{{(x - 3)({x^2} - 4x + 3)}}{{(x - 3)({x^2} - 2{x^2} - 6x + 9)}} = \mathop {\lim }\limits_{x \to 3} \dfrac{{{x^2} - 4x + 3}}{{{x^3} - 2{x^2} - 6x + 9}}
Cancelling out (x - 3) from both the numerator and the denominator
There is still another common (x - 3) left in both the numerator and the denominator.
Factorization again leads to
Cancelling out (x-3) again from the numerator and the denominator.
\mathop {\lim }\limits_{x \to 3} \dfrac{{(x - 1)(x - 3)}}{{(x - 3)({x^2} + x - 3)}} = \mathop {\lim }\limits_{x \to 3} \dfrac{{x - 1}}{{{x^2} + x - 3}} = \dfrac{2}{9}
     Example: 2     

Evaluate the following limits:
(a) \mathop {\lim }\limits_{x \to 4} \dfrac{{\sqrt {1 + 2x}  - 3}}{{\sqrt x  - 2}}
(b) \mathop {\lim }\limits_{x \to 1} \dfrac{{{x^2} - \sqrt x }}{{\sqrt x  - 1}}
(c) \mathop {\lim }\limits_{x \to \infty } \dfrac{{\sqrt {{x^2} + 1}  - \sqrt[3]{{{x^2} + 1}}}}{{\sqrt[4]{{{x^4} + 1}} - \sqrt[5]{{{x^4} + 1}}}}
(d) \mathop {\lim }\limits_{x \to 1} \dfrac{{(x + {x^2} +\ldots+ {x^n}) - n}}{{x - 1}}
Solution: 2-(a)

This limit can evidently be solved by rationalising both the numerator and the denominator.

\mathop {\lim }\limits_{x \to 4} \left\{ {\dfrac{{\sqrt {1 + 2x}  - 3}}{{\sqrt x  - 2}} \times \dfrac{{\sqrt {1 + 2x}  + 3}}{{\sqrt {1 + 2x}  + 3}} \times \dfrac{{\sqrt x  + 2}}{{\sqrt x  + 2}}} \right\}
 = \mathop {\lim }\limits_{x \to 4} \left\{ {\dfrac{{2x - 8}}{{x - 4}} \times \dfrac{{\sqrt x  + 2}}{{\sqrt {1 + 2x}  + 3}}} \right\}
 = 2\mathop {\lim }\limits_{x \to 4} \dfrac{{\sqrt x  + 2}}{{\sqrt {1 + 2x}  + 3}}
 = \dfrac{4}{3}
Solution: 2-(b)

This can be solved by rationalisation again.
\mathop {\lim }\limits_{x \to 1} \dfrac{{{x^2} - \sqrt x }}{{\sqrt x  - 1}} \times \dfrac{{{x^2} + \sqrt x }}{{{x^2} + \sqrt x }} \times \dfrac{{\sqrt x  + 1}}{{\sqrt x  + 1}}  = \mathop {\lim }\limits_{x \to 1} \left\{ {\dfrac{{{x^4} - x}}{{x - 1}} \times \dfrac{{\sqrt x  + 1}}{{{x^2} + \sqrt x }}} \right\}  = \mathop {\lim }\limits_{x \to 1} \left\{ {x\left( {{x^2} + x + 1} \right) \times \dfrac{{\sqrt x  + 1}}{{{x^2} + \sqrt x }}} \right\}
Solution: 2-(c)

This limit is of the indeterminate form \dfrac{{\infty  - \infty }}{{\infty  - \infty }} (and look very complicated !)
However, division of both the numerator and denominator by x directly reduces the limit to a determinate form.
 = \mathop {\lim }\limits_{x \to \infty } \left\{ {\dfrac{{\sqrt {\dfrac{{{x^2} + 1}}{{{x^2}}}}  - \sqrt[3]{{\dfrac{{{x^2} + 1}}{{{x^3}}}}}}}{{\sqrt[4]{{\dfrac{{{x^4} + 1}}{{{x^4}}}}} - \sqrt[5]{{\dfrac{{{x^4} + 1}}{{{x^5}}}}}}}} \right\}
= \mathop {\lim }\limits_{x \to \infty } \left\{ {\dfrac{{\sqrt {1 + \dfrac{1}{{{x^2}}}}  - \sqrt[3]{{\frac{1}{x} + \dfrac{1}{{{x^3}}}}}}}{{\sqrt[4]{{1 + \dfrac{1}{{{x^4}}}}} - \sqrt[5]{{\dfrac{1}{x} + \dfrac{1}{{{x^5}}}}}}}} \right\}
{\sqrt[3]{{\dfrac{1}{x} + \dfrac{1}{{{x^3}}}}}} and {\sqrt[5]{{\dfrac{1}{x} + \dfrac{1}{{{x^5}}}}}} these two terms tend to 0 as x \to \infty
=1
Solution: 2-(d)

since the denominator is x - 1, we can get a hint that the numerator
(x + {x^2} + \ldots + {x^n}) - n
can be written as
(x - 1) + ({x^2} - 1) +\ldots + ({x^n} - 1)
so that
\mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {x + {x^2} + \ldots+ {x^n}} \right) - n}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \left\{ {\dfrac{{x - 1}}{{x - 1}} + \dfrac{{{x^2} - 1}}{{x - 1}} +\ldots + \dfrac{{{x^n} - 1}}{{x - 1}}} \right\}
 = 1 + 2 + 3 + \ldots + n
 = \dfrac{{n\left( {n + 1} \right)}}{2}
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