How to Solve Indeterminate Forms of Limits

Here, you will learn how to solve indeterminate forms of limits and general methods to be used to evaluate limits with examples.

Let’s begin – 

Indeterminate Forms of Limits

\(0\over 0\), \(\infty \over \infty\), \(\infty – \infty\),\(0\times \infty\), \(1^{\infty}\), \(0^0\), \({\infty}^0\)

Note :

(i)  Here 0, 1 are not exact, infact both are approaching to their corresponding values.

(ii)  We cannot plot \(\infty\) on the paper. Infinity is a symbol & not a number. It does not obey the laws of elementary algebra.

General methods to be used to evaluate limits

(a) Factorisation

\(\displaystyle{\lim_{x \to a}}\) \(x^n – a^n\over {x – a}\) = n\(a^{n – 1}\)

Example : Evaluate the limit : \(\displaystyle{\lim_{x \to 2}}\) [\(1\over {x – 2}\) – \(2(2x – 3)\over {x^3 – 3x^2 + 2x}\)]

Solution : We have

\(\displaystyle{\lim_{x \to 2}}\) [\(1\over {x – 2}\) – \(2(2x – 3)\over {x^3 – 3x^2 + 2x}\)] = \(\displaystyle{\lim_{x \to 2}}\) [\(1\over {x – 2}\) – \(2(2x – 3)\over {x(x – 1)(x – 2)}\)]

= \(\displaystyle{\lim_{x \to 2}}\) [\(x^2 – 5x + 6\over {x(x – 1)(x – 2)}\)] = \(\displaystyle{\lim_{x \to 2}}\) [\((x – 2)(x – 3)\over {x(x – 1)(x – 2)}\)] = \(\displaystyle{\lim_{x \to 2}}\) [\(x – 3\over {x(x – 1)}\)] = \(-1\over 2\)

(b) Rationalization or Double Rationalization

Example : Evaluate the limit : \(\displaystyle{\lim_{x \to 1}}\) [\({4 – \sqrt{15x + 1}}\over {2 – \sqrt{3x + 1}}\)]

Solution : \(\displaystyle{\lim_{x \to 1}}\) [\({4 – \sqrt{15x + 1}}\over {2 – \sqrt{3x + 1}}\)]

= \(\displaystyle{\lim_{x \to 1}}\) \({(4 – \sqrt{15x + 1})(2 + \sqrt{3x + 1})(4 + \sqrt{15x + 1})}\over {(2 – \sqrt{3x + 1})(4 + \sqrt{15x + 1})(2 + \sqrt{3x + 1})}\) = \(\displaystyle{\lim_{x \to 1}}\) \((15 – 15x)\over {3 – 3x}\)\(\times\)\(2 + \sqrt{3x + 1}\over {4 + \sqrt{15x + 1}}\) = \(5\over 2\)

(c) Limit at Infinity

(i) Divide by greatest power of x in numerator and denominator

(ii) Put x = \(1\over y\) and apply \(y \to 0\)

Example : Evaluate the limit : \(\displaystyle{\lim_{x \to \infty}}\) \(x^2 + x + 1\over {3x^2 + 2x – 5}\)

Solution : \(\displaystyle{\lim_{x \to \infty}}\)\(x^2 + x + 1\over {3x^2 + 2x – 5}\)       (\(\infty\over \infty\) form)

Put x = \(1\over y\)

Limit = \(\displaystyle{\lim_{y \to 0}}\) \(1 + y + y^2\over {3 + 2y – 5y^2}\) = \(1\over 3\)

Hope you learnt how to solve indeterminate forms of limits and general methods to be used to evaluate limits. To learn more on limits practice more questions and read more examples and get ahead in competition. Good Luck!

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