# Standard Limits – Problems

## 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.