How do I factor this? Simplifying for limit.

Question

How do I factor $(s^5 - a^5)/(s^2 - a^2)$ ? I need to simplify it so I can find the limit as $s \rightarrow a$.

Answer 1

Clearly $s=a$ is a root of the numerator, so $s-a$ is a factor. In fact, $$s^5-a^5 = (s-a)(s^4 + s^3a + s^2 a^2 + s a^3 + a^4)$$ as can be verified by carrying out the multiplication on the right hand side. Therefore, we can reduce the fraction to $$\frac{s^4 + s^3a + s^2 a^2 + s a^3 + a^4}{s+a}$$

Answer 2

You can use the following for any positive integer $n$ : $$A^n-B^n=(A-B)(A^{n-1}+A^{n-2}B+\cdots+AB^{n-2}+B^{n-1}).$$ Hence, you'll get $$\frac{s^5-a^5}{s^2-a^2}=\frac{\color{red}{(s-a)}(s^4+s^3a+s^2a^2+sa^3+a^4)}{\color{red}{(s-a)}(s+a)}=\frac{s^4+s^3a+s^2a^2+sa^3+a^4}{s+a}.$$

Answer 3

$$\frac{s^5-a^5}{s^2-a^2}$$
$$=\frac{(s-a)(s^4+s^3a+s^2a^2+sa^3+a^4)}{(s-a)(s+a)}$$
$$=\frac{(s^4+s^3a+s^2a^2+sa^3+a^4)}{(s+a)}$$

Answer 4

Assuming that the problem is not limited to the calculation of the limit, you could simplify even more your expression defining $s=t \times a$. So $$\frac{s^5-a^5}{s^2-a^2}=\frac{a^5 t^5-a^5}{a^2t^2-a^2}=a^3\frac{ t^5-1}{t^2-1}=$$ $t=1$ cancels both numerator and denominator; so, as explained in answers and comments,$$\frac{s^5-a^5}{s^2-a^2}=a^3 \frac{(t-1)(t^4+t^3+t^2+t+1)}{(t-1)(t+1)}=a^3 \frac{t^4+t^3+t^2+t+1}{t+1}$$ It could be shown that the numerator does not show any real root by only complex roots; so let us write $$t^4+t^3+t^2+t+1=(t^2+At+1)(t^2+Bt+1)$$ Developing the rhs and identifying the coefficients of the same power leads to two equations $A+B=1$ and $A\times B=-1$ thet is to say the sum and the product of the roots of a quadratic equation; solving, we find $$A=\frac{1}{2} \left(1-\sqrt{5}\right)$$ $$B=\frac{1}{2} \left(1+\sqrt{5}\right)$$ So, as a final result $$\frac{s^5-a^5}{s^2-a^2}=a^3\frac{(t^2+At+1)(t^2+Bt+1)}{t+1}$$ rembering that $t=\frac{s}{a}$

Answer 5

I have normally seen people using standard limits like $$\lim_{x \to 0}\frac{\sin x}{x} = 1, \lim_{x \to 0}\frac{\log(1 + x)}{x} = 1,\lim_{x \to 0}\frac{e^{x} - 1}{x} = 1$$ for trigonometric, logarithmic and exponential limit problems, but rarely do I notice the use of the standard limit $$\lim_{x \to a}\frac{x^{n} - a^{n}}{x - a} = na^{n - 1}\tag {1}$$ for algebraical limits.

In general if the algebraical simplification is small like finding an easy factor or rationalizing some square roots, it is preferred over the use of algebraical limit formula $(1)$ above. But in case you find the factorization or rationalization of radicals difficult it is better to use the formula $(1)$. Thus we have $$\begin{aligned}L &= \lim_{s \to a}\frac{s^{5} - a^{5}}{s^{2} - a^{2}}\\ &= \lim_{s \to a}\dfrac{\dfrac{s^{5} - a^{5}}{s - a}}{\dfrac{s^{2} - a^{2}}{s - a}}\\ &= \frac{5a^{4}}{2a} = \frac{5}{2}a^{3}\end{aligned}$$ Note that the derivation requires $a \neq 0$. But when $a = 0$ the result is easily seen to be $0$ and hence the answer $5a^{3}/2$ remains valid in this case too.