Examples On Evaluating Limits Set-1

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Example - 1

Evaluate the following limits

\(\begin{align}&{\rm\bf{(a)}}\quad \mathop {\lim }\limits_{x \to 2} \left( {\frac{2}{{x(x - 2)}} - \frac{1}{{{x^2} - 3x + 2}}} \right) \qquad \qquad
{\rm\bf{(b)}}\quad \mathop {\lim }\limits_{n \to \infty } \left[ {(1 + x)(1 + {x^2})(1 + {x^4})....(1 + {x^{{2^n}}})} \right]\,\,\,|x|\, < 1\\
&{\rm\bf{(c)}} \quad \mathop {\lim }\limits_{x \to 3} \,\,\,\frac{{{x^3} - 7{x^2} + 15x - 9}}{{{x^4} - 5{x^2} + 27x - 27}}\end{align}\)

Solution: (a) This limit is of the indeterminate form \(\infty  - \infty \) . Combining the two fractions in this limit should lead to a cancellation of the factor giving rise to this indeterminacy, i.e. (x – 2)

\[\mathop {\lim }\limits_{x \to 2} \,\left( {\frac{2}{{x(x - 2)}} - \frac{1}{{{x^2} - 3x + 2}}} \right)\,\,\]

(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 \(\begin{align}\frac{{1 - x}}{{1 - x}}.\end{align}\) This is what happens:

Since \(|x|\,\, < 1,\;\mathop {\lim }\limits_{n \to \infty } {x^{{2^{n + 1}}}} = 0\)

Hence, the value of the limit is \(\frac{1}{{1 - x}}\)

(c) The numerator and denominator both tend to 0 as \(x \to 3\) because of the common factor (x – 3).

Hence, factorization leads to :

There is still another common (x – 3) left in both the numerator and the denominator.

Factorization again leads to

Example - 2

Evaluate the following limits:

\[\;(a)\quad \mathop {\lim }\limits_{x \to 4} \frac{{\sqrt {1 + 2x}  - 3}}{{\sqrt x  - 2}}\] \[\;(b)\quad \mathop {\lim }\limits_{x \to 1} {\mkern 1mu} {\mkern 1mu} \frac{{{x^2} - \sqrt x }}{{\sqrt x  - 1}}\] \[\;(c)\quad {\rm{}}\mathop {\lim }\limits_{x \to \infty } {\mkern 1mu} {\mkern 1mu} \frac{{\sqrt {{x^2} + 1}  - \sqrt[3]{{{x^2} + 1}}}}{{\sqrt[4]{{{x^4} + 1}} - \sqrt[5]{{{x^4} + 1}}}}\] \[\;(d)\quad \mathop {\lim }\limits_{x \to 1} {\mkern 1mu} {\mkern 1mu} \frac{{(x + {x^2} + ...... + {x^n}) - n}}{{x - 1}}\]

Solution: (a) This limit can evidently be solved by rationalising both the numerator and the denominator.

\[\begin{align} &= \mathop {\lim }\limits_{x \to 4} \left\{ {\frac{{2x - 8}}{{x - 4}}\,\,\,\, \times \,\,\,\,\frac{{\sqrt x + 2}}{{\sqrt {1 + 2x} + 3}}} \right\} \\ &= 2\mathop {\lim }\limits_{x \to 4} \frac{{\sqrt x + 2}}{{\sqrt {1 + 2x} + 3}} \\ &= \frac{4}{3} \\ \end{align} \]

(b) This can be solved by rationalisation again.

\[\begin{align} &= \mathop {\lim }\limits_{x \to 1} \left\{ {\frac{{{x^4} - x}}{{x - 1}}\,\,\,\, \times \,\,\,\,\,\frac{{\sqrt x + 1}}{{{x^2} + \sqrt x }}} \right\} \\& = \mathop {\lim }\limits_{x \to 1} \left\{ {x\left( {{x^2} + x + 1} \right)\,\,\,\, \times \,\,\,\,\frac{{\sqrt x + 1}}{{{x^2} + \sqrt x }}} \right\} \\ &= 3\\ \end{align} \]

(c) This limit is of the indeterminate form \(\frac{{\infty - \infty }}{{\infty - \infty }}\) (and look very complicated !)

However, division of both the numerator and denominator by x directly reduces the limit to a determinate form.

(d) since the denominator is x–1, we can get a hint that the numerator

\[(x{\rm{ \;}} + {\rm{\; }}{x^2} + {\rm{\; }}... + {x^n}){\rm{ \;}}-{\rm{\; }}n\]

can be written as

\[\left( {x{\rm{\;}}-{\rm{\;}}1} \right){\rm{\;}} + {\rm{ \;}}\left( {{x^2}-{\rm{\; }}1} \right){\rm{\;}} + {\rm{\; }}...{\rm{\;}} + {\rm{\;}}\left( {{x^n}-{\rm{ }}1} \right)\]

so that

\[\begin{align}\mathop {\lim }\limits_{x \to 1} \frac{{\left( {x + {x^2} + ... + {x^n}} \right) - n}}{{x - 1}} &= \mathop {\lim }\limits_{x \to 1} \left\{ {\frac{{x - 1}}{{x - 1}} + \frac{{{x^2} - 1}}{{x - 1}} + ... + \frac{{{x^n} - 1}}{{x - 1}}} \right\}\\&
 = {\rm{\;}}1{\rm{\;}} + {\rm{\;}}2{\rm{\;}} + {\rm{\;}}3{\rm{\;}} + {\rm{\;}}... + {\rm{\;}}n\\
 &= \frac{{n\left( {n + 1} \right)}}{2}\end{align}\]