Evaluate $\lim\limits_{x\to 1}\frac{x^{10}-1}{\sqrt x-1}$ without L'Hopital's rule

Question

I was faced with a problem about evaluating limits.

I know for a fact that when a limit is indeterminate when substituted with the value $x$ is approaching, I differentiate by L'Hopital's Rule. But then my professor said that it was for higher Math that I should "manually" evaluate the limits. Is there any step-by-step procedure on how to do this? An example is cited below: $$\lim\limits_{x\to 1}\frac{x^{10}-1}{\sqrt x-1}.$$

By the use of differentiation, I got $20$ as the answer. But how can I show the answer without differentiation? I tried rationalizing the denominator but it ends up with $0$ as the answer. I have also tried to factor the denominator after rationalizing by $(x)^3-(1)^3$ but still it's the same dead end.

Answer 1

Hints:

(1) $$x^{10} - 1 = (x - 1)(x^9 + x^8 + x^7 + x^6 + x^5 + x^4 + x^3 + x^2 + x + 1)$$

(2) $$\sqrt{x} - 1 = \frac{x - 1}{\sqrt{x} + 1}$$

Can you take it from here?

Answer 2

You should cancel out the terms to get rid of $\frac{0}{0}$ situation. Observe that \begin{equation} x^{10}-1=(x-1)(x^9+\dots+1) \end{equation} Hence \begin{equation} \frac{x^{10}-1}{\sqrt{x}-1}=\frac{(x-1)}{\sqrt{x}-1}(x^9+\dots+1)=)(\sqrt{x}+1)(x^9+\dots+1) \end{equation} The right hand side is now simple to evaluate as $x\rightarrow 1$ in particular it is equal to $20$.

Answer 3

$\lim_{x\to1} \frac{x^{10}-1}{\sqrt{x}-1} = \lim_{x\to1} \frac{\sqrt{x}^{20}-1}{\sqrt{x}-1}= \lim_{x\to 1} (\sqrt{x}^{19}+\sqrt{x}^{18}+...+\sqrt{x}+1 )= 20$

Answer 4

Alternative: Set $u=\sqrt x-1$, then $$ x^{10}-1=(1+u)^{20}-1=1+20u+\sum_{k=2}^{20}\binom{20}k u^k -1 \\=u·\left(20+u·\sum_{k=2}^{20}\binom{20}k u^{k-2}\right) $$ which avoids collecting the 20 limits to 1 in the geometric sum formula.

Answer 5

Set $u=\sqrt x$. Then $$\frac{x^{10}-1}{\sqrt x-1}=\frac{u^{20}-1}{u-1}\to(u^{20})'\Big\vert_{u=1}=20.$$