Question

\[\lim_{x \to a} \frac{x^n - a^n}{x - a}\]  is equal at 

Options

• naⁿ 

• naⁿ⁻¹ 

• na 

•  1

   

Answer 1

naⁿ⁻¹ 

^{\[\lim_{x \to a} \frac{x^n - a^n}{x - a}\]
\[ = \lim_{x \to a^+} \frac{x^n - a^n}{x - a} \left[ \because f\left( x \right) exists, \lim_{x \to a} f\left( x \right) = \lim_{x \to a^+} f\left( x \right) \right]\]
\[ = \lim_{h \to 0} \frac{\left( a + h \right)^n - a^n}{a + h - a}\]
\[ = \lim_{h \to 0} a^n \frac{\left[ \left( 1 + \frac{h}{a} \right)^n - 1 \right]}{h}\]
\[ = a^n \lim_{h \to 0} \left[ 1 + n \cdot \frac{h}{a} + \frac{n\left( n - 1 \right)}{2!}\frac{h^2}{a^2} . . . + . . . - 1 \right]\]
\[ = a^n \lim_{h \to 0} \left[ \frac{n}{a} + \frac{h\left( h - 1 \right)}{2!} \frac{h}{a^2} + . . . \right]\]
\[ = a^n \frac{n}{a}\]
\[ = n a^{n - 1}\]}