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