# Limits – II

## Limits – II

Evaluation of Algebraic limits using some standard limits: We know that binomial expansion for any rational power.

$${{(1+x)}^{n}}=1+nx+\frac{n(n-1)}{2!}{{x}^{2}}+\frac{n(n-1)(n-2)}{3!}{{x}^{3}}+….$$ .

Where |x| < 1

1. Theorem: If n ϵ Q, then  $$\underset{x\to a}{\mathop{\lim }}\,\frac{{{x}^{n}}-{{a}^{n}}}{x-a}=n.{{a}^{n-1}}$$.

Proof: We have

$$\underset{x\to a}{\mathop{\lim }}\,\frac{{{x}^{n}}-{{a}^{n}}}{x-a}$$.

$$\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,\frac{{{x}^{n}}-{{a}^{n}}}{x-a}$$.

$$\underset{h\to 0}{\mathop{\lim }}\,\frac{{{(a+h)}^{n}}-{{a}^{n}}}{a+h-a}$$.

$$\underset{h\to 0}{\mathop{\lim }}\,\frac{{{a}^{n}}\{{{(1+h/a)}^{n}}-1\}}{h}$$.

$${{a}^{n}}\underset{x\to 0}{\mathop{\lim }}\,\frac{\left( 1+n.\frac{h}{a} \right)-1}{h}$$.

(when x → 0, (1 + x) ⁿ = 1 + n x)

= aⁿ. n/a

= n. aⁿ⁻¹.

Hence proved

$$\underset{x\to a}{\mathop{\lim }}\,\frac{{{x}^{n}}-{{a}^{n}}}{x-a}=n.{{a}^{n-1}}$$.

2. Theorem: $$\underset{x\to 0}{\mathop{\lim }}\,\frac{{{(1+x)}^{n}}-1}{x}=n$$.

Proof: We have

= $$\underset{x\to 0}{\mathop{\lim }}\,\frac{{{(1+x)}^{n}}-1}{x}$$.

$${{(1+x)}^{n}}=1+nx+\frac{n(n-1)}{2!}{{x}^{2}}+\frac{n(n-1)(n-2)}{3!}{{x}^{3}}+….$$.

= $$\underset{x\to 0}{\mathop{\lim }}\,\frac{1+nx+\frac{n(n-1)}{2}{{x}^{2}}+……..-1}{x}$$.

= $$\underset{x\to 0}{\mathop{\lim }}\,\frac{nx+\frac{n(n-1)}{2}{{x}^{2}}+……..}{x}$$.

= $$\underset{x\to 0}{\mathop{\lim }}\,n+\frac{n(n-1)}{2}x+…..$$.

= n

Hence proved  $$\underset{x\to 0}{\mathop{\lim }}\,\frac{{{(1+x)}^{n}}-1}{x}=n$$.