Hey limit, do you exist?

It's a limit with form $\frac{0}{0}$ undetermined. $\displaystyle \lim_{x \to 1} \frac{x^{13} - 1}{ x^8 - 1} = \lim_{x \to 1} \frac{(x - 1)(x^{12} + x^{11} + x^{10} + ... + x + 1)}{(x - 1)(x^7 + x^6 + x^5 +.... + x + 1)}$ $= \displaystyle \lim_{x \to 1} \frac{x^{12} + x^{11} + x^{10} + ... + x + 1}{x^7 + x^6 + x^5 +.... + x + 1} =$ Now, substituing at $x = 1$ $= \frac{13}{8} = 1.625$

It is visible that $\displaystyle 1$ is a root of the polynomial $\displaystyle({ x }^{ 13 }-1)$ . Then clearly the polynomial has $\displaystyle(x-1)$ as one of its factors. Similarly $\displaystyle({ x }^{ 8 }-1)$ also has $\displaystyle(x-1)$ as one of its factors.

Therefore, by using algebraic division to obtain the other factors, the given polynomial function becomes:

$\displaystyle \frac { (x-1)({ x }^{ 12 }+{ x }^{ 11 }+{ x }^{ 10 }+...{ x }^{ 2 }+x+1) }{ (x-1)(x+1)({ x }^{ 2 }+1)({ x }^{ 4 }+1) }$

= $\displaystyle \frac { ({ x }^{ 12 }+{ x }^{ 11 }+{ x }^{ 10 }+...{ x }^{ 2 }+x+1) }{ (x+1)({ x }^{ 2 }+1)({ x }^{ 4 }+1) }$

Here, we can substitute $\displaystyle x=1$ such that the limit doesn't stand out to be undefined. Hence, by substituting we have the limit of the function as $\displaystyle \boxed{\frac{13}{8}}$ , while $\displaystyle x$ tends to $\displaystyle 1$ .

Use L-Hospital's rule to evaluate the limit because you first get 0/0 which is indeterminate. So, evaluate limit as x approaches 1 of 13x^12 / 8x^7 , which is 13/8