Standard Limits – Problems
Theorem 1: If n is a rational number and a > 0 then \(\underset{x\to a}{\mathop{\lim }}\,\frac{{{x}^{n}}-{{a}^{n}}}{x-a}=n.{{a}^{n-1}}\).
If n is positive integer and a > 0 then for any a ϵ R, \(\underset{x\to a}{\mathop{\lim }}\,\frac{{{x}^{n}}-{{a}^{n}}}{x-a}=n.{{a}^{n-1}}\).
If n is real number and a > 0 then \(\underset{x\to a}{\mathop{\lim }}\,\frac{{{x}^{n}}-{{a}^{n}}}{x-a}=n.{{a}^{n-1}}\).
If m and n are any real numbers and a > 0, then \(\underset{x\to a}{\mathop{\lim }}\,\frac{{{x}^{m}}-{{a}^{n}}}{x-a}=\frac{m}{n}.{{a}^{m-n}}\).
Theorem 2: If 0 < x < π/2 then sinx < x < tanx.
If – π/2 < x < 0, then the tanx < x < sinx.
If 0 < |x| < π/2 then |sinx| < |x| < |tanx|.
Example 1: \(\underset{x\to a}{\mathop{\lim }}\,\frac{\tan (x-a)}{{{x}^{2}}-{{a}^{2}}}\) (a ≠ 0).
Solution: Given that,
\(\underset{x\to a}{\mathop{\lim }}\,\frac{\tan (x-a)}{{{x}^{2}}-{{a}^{2}}}\) (a ≠ 0),
= \(\underset{x\to a}{\mathop{\lim }}\,\frac{\tan (x-a)}{(x-a)(x+a)}\),
= \(\underset{x\to a}{\mathop{\lim }}\,\frac{\tan (x-a)}{(x-a)}\ .\underset{x\to a}{\mathop{\lim }}\,\frac{1}{x+a}\),
In the first limit put x – a = h so that as x → a, h → 0.
= \(\underset{x\to a}{\mathop{\lim }}\,\frac{\tanh }{h}\ .\frac{1}{a+a}\),
= 1. ½ a
= ½ a
Example 2: \(\underset{x\to 0}{\mathop{\lim }}\,\frac{1-\cos 2mx}{{{\sin }^{2}}nx}\) (m, n ϵ -2).
Solution: Given that,
\(\underset{x\to 0}{\mathop{\lim }}\,\frac{1-\cos 2mx}{{{\sin }^{2}}nx}\) (m, n ϵ -2),
\(=\underset{x\to 0}{\mathop{\lim }}\,\frac{2{{\sin }^{2}}mx}{{{\sin }^{2}}nx}\),
\(=\underset{x\to 0}{\mathop{\lim }}\,\frac{2{{\sin }^{2}}mx}{{{\sin }^{2}}nx}\times \frac{{{x}^{2}}}{{{x}^{2}}}\),
\(=\underset{x\to 0}{\mathop{\lim }}\,\frac{\frac{2{{\sin}^{2}}mx}{{{x}^{2}}}}{\frac{{{\sin }^{2}}nx}{{{x}^{2}}}}\),
= 2m²/n².