{"id":3950,"date":"2021-08-11T11:59:21","date_gmt":"2021-08-11T11:59:21","guid":{"rendered":"https:\/\/mathemerize.com\/?p=3950"},"modified":"2022-02-03T18:28:30","modified_gmt":"2022-02-03T12:58:30","slug":"indeterminate-forms-of-limits","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/indeterminate-forms-of-limits\/","title":{"rendered":"How to Solve Indeterminate Forms of Limits"},"content":{"rendered":"
Here, you will learn how to solve indeterminate forms of limits and general methods to be used to evaluate limits with examples.<\/p>\n
Let’s begin – <\/p>\n
\\(0\\over 0\\), \\(\\infty \\over \\infty\\), \\(\\infty – \\infty\\),\\(0\\times \\infty\\), \\(1^{\\infty}\\), \\(0^0\\), \\({\\infty}^0\\)<\/p>\n
Note :<\/strong><\/p>\n (i) Here 0, 1 are not exact, infact both are approaching to their corresponding values.<\/p>\n (ii) We cannot plot \\(\\infty\\) on the paper. Infinity is a symbol & not a number. It does not obey the laws of elementary algebra.<\/p>\n \\(\\displaystyle{\\lim_{x \\to a}}\\) \\(x^n – a^n\\over {x – a}\\) = n\\(a^{n – 1}\\)<\/p>\n\n\n Example : <\/span>Evaluate the limit : \\(\\displaystyle{\\lim_{x \\to 2}}\\) [\\(1\\over {x – 2}\\) – \\(2(2x – 3)\\over {x^3 – 3x^2 + 2x}\\)] <\/p>\n Solution : <\/span>We have Example : <\/span>Evaluate the limit : \\(\\displaystyle{\\lim_{x \\to 1}}\\) [\\({4 – \\sqrt{15x + 1}}\\over {2 – \\sqrt{3x + 1}}\\)] <\/p>\n Solution : <\/span>\n \\(\\displaystyle{\\lim_{x \\to 1}}\\) [\\({4 – \\sqrt{15x + 1}}\\over {2 – \\sqrt{3x + 1}}\\)] (i) Divide by greatest power of x in numerator and denominator<\/p>\n (ii) Put x = \\(1\\over y\\) and apply \\(y \\to 0\\)<\/p>\n\n\n Example : <\/span>Evaluate the limit : \\(\\displaystyle{\\lim_{x \\to \\infty}}\\) \\(x^2 + x + 1\\over {3x^2 + 2x – 5}\\) <\/p>\n Solution : <\/span>\n \\(\\displaystyle{\\lim_{x \\to \\infty}}\\)\\(x^2 + x + 1\\over {3x^2 + 2x – 5}\\) (\\(\\infty\\over \\infty\\) form) 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!<\/p>\n\n\nGeneral methods to be used to evaluate limits<\/h2>\n
(a) Factorisation<\/strong><\/h4>\n
\n\t\t \\(\\displaystyle{\\lim_{x \\to 2}}\\) [\\(1\\over {x – 2}\\) – \\(2(2x – 3)\\over {x^3 – 3x^2 + 2x}\\)] = \n\t\t \\(\\displaystyle{\\lim_{x \\to 2}}\\) [\\(1\\over {x – 2}\\) – \\(2(2x – 3)\\over {x(x – 1)(x – 2)}\\)]
\n\t\t = \\(\\displaystyle{\\lim_{x \\to 2}}\\) [\\(x^2 – 5x + 6\\over {x(x – 1)(x – 2)}\\)] = \n\t\t \\(\\displaystyle{\\lim_{x \\to 2}}\\) [\\((x – 2)(x – 3)\\over {x(x – 1)(x – 2)}\\)] = \n\t\t \\(\\displaystyle{\\lim_{x \\to 2}}\\) [\\(x – 3\\over {x(x – 1)}\\)] = \\(-1\\over 2\\)
<\/p>\n\n\n(b) Rationalization or Double Rationalization<\/strong><\/h4>\n\n\n
\n\t\t = \\(\\displaystyle{\\lim_{x \\to 1}}\\) \\({(4 – \\sqrt{15x + 1})(2 + \\sqrt{3x + 1})(4 + \\sqrt{15x + 1})}\\over \n\t\t {(2 – \\sqrt{3x + 1})(4 + \\sqrt{15x + 1})(2 + \\sqrt{3x + 1})}\\)\n\t\t = \\(\\displaystyle{\\lim_{x \\to 1}}\\) \\((15 – 15x)\\over {3 – 3x}\\)\\(\\times\\)\\(2 + \\sqrt{3x + 1}\\over {4 + \\sqrt{15x + 1}}\\)\n\t\t = \\(5\\over 2\\)
<\/p>\n\n\n(c) Limit at Infinity<\/strong><\/h4>\n
Put x = \\(1\\over y\\)
Limit = \\(\\displaystyle{\\lim_{y \\to 0}}\\) \\(1 + y + y^2\\over {3 + 2y – 5y^2}\\) = \\(1\\over 3\\)
<\/p>\n\n\n