# Limit problem - help please guys

• March 4th 2008, 07:50 AM
janvdl
Limit problem - help please guys
Hey guys,

I've got a big test on friday and I was working out a past paper.

I would appreciate it if you could show all the steps.

The question is as follows:

$\mathop {\lim }\limits_{x \to 1} \frac{{x^{17} - 1}}
{{x - 1}}
$

There is a tip saying: "Think of a derivative".

And the memo only says:

$f'(1) = 17$

I have no idea how they got that answer.

• March 4th 2008, 08:04 AM
colby2152
You may imply thinking of derivatives as applying L'Hopital's rule.
• March 4th 2008, 08:08 AM
Peritus
you could also notice that:

$
\frac{{x^{17} - 1}}
{{x - 1}} = x^{16} + x^{15} + x^{14} + \cdots + x^2 + x + 1
$

and it's easy to see that the resulting function equals 17 when x approaches 1.
• March 4th 2008, 08:09 AM
PaulRS
Well there are many ways.

I suppose that $f(x)=x^{17}-1$

Right?

That might suggest applying L'Hôpital's rule .

Another Idea would be: $\frac{x^{17}-1}{x-1}=1+x+...+x^{16}$ for all reals $x$ except for 0

So: $\lim_{x\rightarrow{1}}\frac{x^{17}-1}{x-1}=\lim_{x\rightarrow{1}}{\left(1+x+...+x^{16}\rig ht)}$

Or you can consider the change of variable $e^{u}=x$ so $x\rightarrow{1}, u\rightarrow{0}$
• March 4th 2008, 08:12 AM
janvdl
Quote:

Originally Posted by PaulRS
I suppose that $f(x)=x^{17}-1$

Right?

$\frac{x^{17}-1}{x - 1}$
• March 4th 2008, 08:26 AM
wingless
Maybe he tried to say, $f(x) = x^{17}-1$ and $g(x) = x-1$. Then was going to use,
If $\frac{f(a)}{g(a)}$ is in indeterminate form $\left ( \frac{0}{0}, \frac{\infty}{\infty} \right )$
then
$\lim_{x\to a}\frac{f(x)}{g(x)} = \lim_{x\to a}\frac{f'(x)}{g'(x)}$
which is the description of l'hopital's rule.
• March 4th 2008, 08:31 AM
janvdl
I get it now. Thanks for all the help and replies guys, truly appreciated! (Handshake) (Wink)
• March 4th 2008, 08:34 AM
colby2152
Quote:

Originally Posted by janvdl
I get it now. Thanks for all the help and replies guys, truly appreciated! (Handshake) (Wink)

You are welcome! (Handshake)
• March 4th 2008, 08:53 AM
Krizalid
A related problem.