Can we apply L'Hopital rule to 4/0 form?

[jack1234]

1) Can we apply L'Hopital rule to the 4/0 form?
eg
\lim_{x\rightarrow 0} \frac{x+4}{x^2}
=\lim_{x\rightarrow 0} \frac{1}{2x}
=0

2) we know that 0/0 is indefinite form, but is 4/0 indefinite form?

 

[quasar987]

Did you mean to give

\lim_{x\rightarrow 0}\frac{x+4}{x^2}

as an exemple? (click the equation to see the code I used to write it)

Anyway, no when the numerator goes to something finite but the denominator goes to 0, we write that the limit is \infty.

There are seven indeterminate forms, and they are listed here:
http://mathworld.wolfram.com/Indeterminate.html
(scroll down a little bit)

 

[jack1234]

Yes, this is exactly what I mean. Thanks for helping me for the Latex, I am still looking for a manual for it.

For my question, am I right that the answer for both question is
1)No, the answer is \infty
2)No
according to my understanding on your comment?

 

[quasar987]

Yes, you interpreted my answer correctly.

Btw, the convention is to call those kind of limits indeterminate forms and not indefinite forms.

To learn later, I recommend browsing this thread: https://www.physicsforums.com/showthread.php?t=8997

 

[jack1234]

Hi, I have a question here

https://www.physicsforums.com/showthread.php?t=126284
\lim_{x\rightarrow 2} (\frac{|x-3|-|3x-5|}{x^2-5x+6})

So, it is -10/0 form,
is the answer \infty again? Although I think it is not, as I use a the computer to graph it, it is 4.

How do we determine the answer is \infty?

 

[matt grime]

Put x=2 into that expression and what do you get? not -10/0, that is for sure.

If you ever by naive substitution get a/0 where a is not zero then the limit is undefined (in the reals) or infinity for short hand. But you are not in that situation here.

However, around x=2, x-3 is negative and 3x-5 is positive, so near to 2 the expression is the same as (3-x -3x +5)/(x^2-5x+6), and you can simplify that expression by simple algebra.

Don't forget whenever youi hve absolute value signs in that you can (and should) split things up into regions where the expressions inside the abs value signs are positive or negative.

 

[jack1234]

Thanks, I have really calculated it wrongly.