# Limit problem

• Jun 22nd 2013, 09:12 AM
hacker804
Limit problem
So i started calculus and am currently stuck at the following problem.

Q.Evaluate by algebraic techniques:
$\displaystyle \lim_{x\rightarrow a}\frac{x^{n}-a^{n}}{x^{m}-a^{m}}$

The answer provided by the book is:
$\displaystyle \frac{n}{m}x^{n-m}$

Thanks.
• Jun 22nd 2013, 09:44 AM
MINOANMAN
Re: Limit problem

factorise both terms of the fraction EX: x^n-a^n=(x-a)(x^(n-1)+ax^(n-2)+...+a^(n-1)....
do the same for the denominator ..simplify and then the limit when x goes to a is the one you mentioned ....
• Jun 22nd 2013, 11:07 AM
Plato
Re: Limit problem
Quote:

Originally Posted by hacker804
So i started calculus and am currently stuck at the following problem.

Q.Evaluate by algebraic techniques:
$\displaystyle \lim_{x\rightarrow a}\frac{x^{n}-a^{n}}{x^{m}-a^{m}}$

The answer provided by the book is:
$\displaystyle \color{red}\frac{n}{m}x^{n-m}$

Reply #2 is correct in approach, but there is a slight error in the OP.

The correct answer should be $\displaystyle \frac{n}{m}a^{n-m}$
• Jun 22nd 2013, 05:36 PM
hacker804
Re: Limit problem
Yes it's a mistake in the book.Thanks for the help.
• Jun 22nd 2013, 09:24 PM
Prove It
Re: Limit problem
Quote:

Originally Posted by hacker804
So i started calculus and am currently stuck at the following problem.

Q.Evaluate by algebraic techniques:
$\displaystyle \lim_{x\rightarrow a}\frac{x^{n}-a^{n}}{x^{m}-a^{m}}$

The answer provided by the book is:
$\displaystyle \frac{n}{m}x^{n-m}$

Thanks.

Since the limit is of the indeterminate form \displaystyle \displaystyle \begin{align*} \frac{0}{0} \end{align*}, L'Hospital's Rule can be used.

\displaystyle \displaystyle \begin{align*} \lim_{x \to a} \frac{x^n - a^n}{x^m - a^m} &= \lim_{x \to a}\frac{ \frac{d}{dx} \left( x^n - a^n \right) }{ \frac{d}{dx} \left( x^m - a^m \right) } \\ &= \lim_{ x \to a} { \frac{ n\,x^{n - 1} }{ m \, x^{ m - 1} } } \\ &= \lim_{x \to a} \frac{n}{m}\, x^{ n - m} \\ &= \frac{n}{m}\, a^{n - m} \end{align*}