L hoptals rule

Cilva7

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I am having trouble understanding this rule for the question Lim as x goes to 0 of 1/x squared. After taking the first derivative I get 1/2x and then 1/2, but the answer is infinity. Can you explain this algebraically using L hoptals rule?
 
Before you can apply L'Hopital's rule, you have to check whether it applies. Under what conditions can you use it? Can you, here?

Also, please state the instructions given for this problem. I recently helped a student in person who had not noticed that it said to use L'Hopital's rule if it applied and was useful. That led to a good discussion of what those two words meant! Not all the problems needed, or benefited from, the rule, and it wasn't always applicable at all.

As an aside, you might write the problem as "lim{x->0} 1/x^2", if you don't want to try the more complicated formats.
 
L'hopital rule say that when it is applicable you compute the derivative of the numerator over the derivative of the denominator. Can you please explain how you got to went from \(\displaystyle \dfrac{1}{x^2} to \dfrac{1}{2x}?\) You do know that the derivative of 1 is 0?

More importantly, in your example can you use L'hopital's rule?
 
I am having trouble understanding this rule for the question Lim as x goes to 0 of 1/x squared. After taking the first derivative I get 1/2x and then 1/2, but the answer is infinity. Can you explain this algebraically using L hoptals rule?
First I hope you will write Prof Peterson's advice on your the part if your brain that deals with limits.
Here is a different take that is equally useful. The question \(\displaystyle \mathop {\lim }\limits_{x \to 0} \dfrac{1}{{{x^2}}}\) is simply asking you what is the value of the expression \(\displaystyle \dfrac{1}{x^2}\) if \(\displaystyle x\approx 0~?\)
As with most things mathematical there are restrictions on the notation \(\displaystyle x\approx 0\). It means that \(\displaystyle x\) is near zero but not equal to zero. I beg my students to think of that distinction in every limit problem. So if \(\displaystyle x=-0.0003\) then I think all might agree that close to zero.
Now \(\displaystyle x^2=9\cdot 10^{-8}\). Moreover \(\displaystyle \frac{1}{x^2}=\frac{10^8}{9}\) which is a very large number.
Can you see that the limit is infinite?
 
I made a mistake with taking the derivative of one but Isn't this still considered an indeterminate form because it's 0 over 0 when you take the derivative of 0/2 which is the 3rd derivative of the function which then means you can apply L hoptals rule resulting in 1/2?
 
I made a mistake with taking the derivative of one but Isn't this still considered an indeterminate form because it's 0 over 0 when you take the derivative of 0/2 which is the 3rd derivative of the function which then means you can apply L hoptals rule resulting in 1/2?
When you plug in 0 in your original limit, do you get an indeterminate form? If not, then it does not matter if you get an indeterminate form after applying L'hopital's rule. Why? Because you were NOT allowed to apply L'hopital rule in the 1st place.

Also, let's work with what you got. You got 0/2. Sure you get 0/0 if you take the derivative of the top and divide it by the derivative of the bottom. BUT you can't apply L'hopital rule to lim{x-->0) 0/2 because you are not getting an indeterminate form! If you replace x with 0 in 0/2 you get 0/2!, not 0/0. Besides you really should know lim{x-->0) 0/2 = lim{x-->0) 0 = 0. 0/2 = 0, no limit, no L'hopital rule, no derivative, no anything.

lim{x-->0) (1/x^2) = 1/0 if you directly substitute and 1/0 is not \(\displaystyle \dfrac{0}{0}\) nor is it \(\displaystyle \dfrac{\infty}{\infty}\). That is it is NOT in indeterminate form so NO L'hopital's rule! Now go back and read pka response in post#5[/tex]

One last comment. Even if you could use L'hopital rule, why would you? You need to just know like you know your name that 1 /(very small number)^2 is a very large number. How much is 1/(1/100,000)? That is in the form of 1 over a very small number!
 
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