$$\lim_{x \to 1} \frac{x^3 - 1}{x - 1}$$
As $x$ approaches to $1$, if I use the substitution method, it will become undefined. Then, I tried to multiply it by its conjugate but I still get undefined answer. How can I solve it?
Asked
Active
Viewed 851 times
2
-
1I've edited your question to include mathjax to say what I think you intended. Please verify that it's the correct expression. – Jan 24 '16 at 05:22
-
Beside the remarkable identity given in answers, what about L'Hospital rule ? – Claude Leibovici Jan 24 '16 at 05:31
-
1Here is a generalization of this question: Prove that $\lim_{x\rightarrow 1}{\frac{x^n-1}{x-1}}=n$ for all integer n without L'Hôpital – Martin Sleziak Jan 24 '16 at 08:03
4 Answers
3
Use the fact that $x^3-1=(x-1)(x^2+x+1)$.
Tim Raczkowski
- 20,046
-
-
So, divide $x-1$ into $x^3-1$. You can do this either by long division or synthetic division. – Tim Raczkowski Jan 24 '16 at 05:33
-
1In general, $x^n-y^n = (x-y)(x^{n-1}+x^{n-2}y+...+xy^{n-2}+y^{n-1})$, if $n>0$. – Maxis Jaisi Jan 24 '16 at 05:47
2
Hint: You will have $$\frac{x^3-1}{x-1}=\frac{(x-1)(x^2+x+1)}{(x-1)}$$ and $x\ne 1$ since $$a^3-b^3=(a-b)(a^2+b^2+ab)$$
Maxis Jaisi
- 1,553
Dr. Sonnhard Graubner
- 95,283
2
The other answers are cleaner and require no unnecessary machinery, but here's another method you can try if you can't spot the factorisation for messier expressions (but should be used with care). When evaluated at $x=1$, you get $\frac{0}{0}$, which is in an indeterminate form. This hints at L'Hospital Rule, which says that if
$$\lim\limits_{x \to a} \frac{f(x)}{g(x)}= \frac{0}{0}$$ then $$\lim\limits_{x \to a} \frac{f(x)}{g(x)}=\lim\limits_{x \to a} \frac{f'(x)}{g'(x)}$$
Note that we arrive at our answer if $\lim\limits_{x \to a} \frac{f'(x)}{g'(x)}$ exists, or else we try again.
In your case, $f(x)=x^3-1$, $\implies$ $f'(x)=3x^2$. Likewise, $g(x)=x-1$, $\implies$ $g'(x)=1$. Can you continue?
Maxis Jaisi
- 1,553
-
I found a question which match this condition and I apply it to the limit $\lim_{x \to 1) (2-(x+3)^1/2)/ (x - 1)$ – user307537 Jan 24 '16 at 06:11
-
-
ohh... I think that the answer in the book is not correct (1/4). After I use calculator to solve it, I get -1/4. Thank you for teaching me this new method – user307537 Jan 24 '16 at 06:24
-
If you have doubts about your textbook's answer, you can always consult WolframAlpha. You're welcome. – Maxis Jaisi Jan 24 '16 at 06:39
-
This method also works if you your limit evaluates at $\frac{\infty}{\infty}$. – Maxis Jaisi Jan 24 '16 at 06:41