How do I use factoring to calculate this limit?

Question

For the life of me, I cannot figure out how to factor this to solve the limit. It's probably very simple, but I need help. $$\lim_{t\to a}\frac{t^5-a^5}{t^3-a^3}$$

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"Use factoring to calculate this limit" limit as $t$ approaches $a$ $\frac{t^5 - a^5}{t^3 - a^3}$

Answer 1

Hint $\ $ Change variables $\, t = a+x,\,$ so $\,x = t-a \to 0\,$ as $\,t\to a.\,$ Hence, applying the Binomial Theorem, then factoring out and cancelling $\,x\,$ from top and bottom, we obtain

$$\dfrac{t^5-a^5}{t^3-a^3}\, =\, \dfrac{(a+x)^5-a^5}{(a+x)^3-a^3} \,=\, \dfrac{5a^4x+x^2f(x)}{3a^2x+x^2g(x)}\,=\, \dfrac{5a^4+x f(x)}{3a^2+xg(x)}\ \to\ \dfrac{5a^2}3\ \ {\rm as}\ \ x\to 0 $$

Remark $\ $ The same method works for any rational function, i.e. if you shift $\,x\,$ so it is a limit as $\,x\to 0,\:$ then cancelling common factors of $\,x\,$ from the numerator and denominator results in a fraction whose numerator or denominator is nonzero at $\,x=0,\,$ so the limit is determinate, i.e. no longer of indeterminate form $\,0/0.\,$ The point of doing the shift is that it is much easier to both recognize and cancel common factors of $\,x\,$ than it is for common factors of $\,t-a.\,$

Answer 2

Helpful fact: $t^n-a^n = (t-a)(t^{n-1}+t^{n-2}a^1+t^{n-3}a^2+...+t^1a^{n-2}+a^{n-1})$