• Mar 20th 2008, 08:38 AM
kbgemini16
Ok this is the problem:

Find the limit as x approaches zero
0^3 +x/x

Ok.. so I know it's zero over zero so it's undefined but how can I further
distribute the problem ? I remember my professor saying that we could take a step further but I don't know how to do it in this case.

Hopefully this is enough information to get MUCH needed help
• Mar 20th 2008, 08:57 AM
colby2152
Quote:

Originally Posted by kbgemini16
Ok this is the problem:

Find the limit as x approaches zero
0^3 +x/x

Ok.. so I know it's zero over zero so it's undefined but how can I further
distribute the problem ? I remember my professor saying that we could take a step further but I don't know how to do it in this case.

Hopefully this is enough information to get MUCH needed help

1) $x^3 + \frac{x}{x}$

2) $\frac{x^3 + x}{x}$

3) something else?
• Mar 20th 2008, 09:00 AM
teuthid
L'Hopital
This problem is a ripe candidate for L'Hopital's Law. Are you allowed to use it for this problem?
• Mar 20th 2008, 09:02 AM
kbgemini16
When a limit doesn't exist
it's #2 on your reply. And once I plugged the zero into the equation I got 0/0. In a similar problem done in class our professor further simplfied the answer by distributing the orignal function then cancelling out the like terms from the denominator. But I am somewhat fuzzy on the rest and don't know how to further simplify the answer. (Worried)
• Mar 20th 2008, 09:06 AM
kbgemini16
Quote:

Originally Posted by teuthid
This problem is a ripe candidate for L'Hopital's Law. Are you allowed to use it for this problem?

Yes, I would assume so... He told us that 0/0 would not suffice as an answer. Please don't laugh but I've never even heard of that rule... I am researching it now (Sadsmile)
• Mar 20th 2008, 09:36 AM
colby2152
Quote:

Originally Posted by kbgemini16
it's #2 on your reply. And once I plugged the zero into the equation I got 0/0. In a similar problem done in class our professor further simplfied the answer by distributing the orignal function then cancelling out the like terms from the denominator. But I am somewhat fuzzy on the rest and don't know how to further simplify the answer. (Worried)

L'Hopital's rule requires you to know derivatives, which you may have not learned yet.

Take a numerical approach to a limit when you are lost...

$f(x)=\frac{x^3+x}{x}$

$f(1)=2$

$f(0.5)=\frac{0.125+0.5}{0.5} \Rightarrow 1.25$

$f(0.1)=\frac{0.001+0.1}{0.1} \Rightarrow \frac{0.101}{0.1} = 1.01$

$f(0.01)=\frac{0.000001+0.01}{0.01} \Rightarrow \frac{0.010001}{0.01} = 1.0001$

It is obviously going to one.

Now, think about it intuitively. What is happening? Very small values that are cubed are comparable to zero. What is left? Just $x/x$ which simplifies to one.

Also, note that you can just simplify the function...

$f(x)=\frac{x^3+x}{x} \Rightarrow x^2+1$ for $x \ne 0$

Limits do not take the value of a function at the limiting value, but merely the value an infinitesimally small distance away from the limiting value, so doing this primitive simplification is allowable since the limit never looks at $f(0)$ which so happens to be undefined.
• Mar 20th 2008, 02:46 PM
mr fantastic
Quote:

Originally Posted by teuthid
This problem is a ripe candidate for L'Hopital's Law. Are you allowed to use it for this problem?

Sorry, but l'Hospital's Rule is the lazy way.

Quote:

Originally Posted by colby2152
[snip]
you can just simplify the function...

$f(x)=\frac{x^3+x}{x} \Rightarrow x^2+1$ for $x \ne 0$

Limits do not take the value of a function at the limiting value, but merely the value an infinitesimally small distance away from the limiting value, so doing this primitive simplification is allowable since the limit never looks at $f(0)$ which so happens to be undefined.

(Yes)