<p>limit
x>2</p>
<p>(3xsquared) times (1/(x-2))</p>
<p>in the first part the x is squared not the whole function and in the 2nd part x-2 is in the denominator so 1 is divided by x-2 (just to clarify any misconseptions)</p>
<p>Can this be done using L'Hospital's Rule.
How would you do it if using his rule</p>
<p>oh and no using the calculator.</p>
<p>You don't need to use L'Hopital's. It's not needed; the limit approaches infinity. No amount of simplifying will change that.</p>
<p>As to your question: you can't apply L'Hopital's to this problem in a meaningful manner. Sure you can multiply by (x-2)/(x-2), but that does nothing to the end result.</p>
<p>okay that rule can only be used if you have what is an indeterminant form (sp?) meaning you have infiniti over infinite, zero over zero, zero to the infiniti, or infiniti to the zero.</p>
<p>and yes that obviously goes to inifiniti because the power of the numerator is greater than the power of the denominator.</p>
<p>Actually it's because the denominator goes to 0. If the top and bottom were non-zero it'd approach an actual value.</p>
<p>this doesnt count as an indeterminate form, its over zero. since 2-2=0</p>
<p>What do you do if you get something non-zero over zero (not infinaty EX. 2/0 or 100/0?)</p>
<p>It's infinity. Think about it.</p>
<p>seiken, pretty sure that comparing the degree of the numerator and denominator only works when doing limits approaching infinity, not to some value (2 in this case).</p>
<p>The limit does not exist. It's negative infinity from the left and positive infinity from the right. Like others have said, L'Hopital's Rule is not needed, since it is not an indeterminate form.</p>
<p>lol i did my initial response based on the problem, then forgot the initial problem and did my second response based on my first one. Its finals week, i havent slept in 2 days.</p>
<p>if this is the question..</p>
<p>Lim<br>
x -> 2</p>
<h2>3(x^2)</h2>
<p>(x-2)</p>
<p>then yes, l'hopital's rule is applicable.
it becomes (6x)/1
at x = 2, this is 12.</p>
<p>if this isn't the question, my bad</p>
<p>oops!!!</p>
<p>sorry my bad.
I forgot that l'hopital's rule requires 0/0, etc.
sorry everyone!</p>
<p>thanks for the help i got it.</p>
<p>What do you do for limits if you get something over 0 but not 0 or infinity on the numerator like 10/0
do u still use l'hospital rule?
Do u factor?</p>
<p>vh? you are not supposed to open old threads. I know this is not too old,, but just saying. </p>
<p>As for your question, l'hopital's rule only applies for 0/0, infinity/infinity, etc. If, like in your example, the limit works out to 10/0, the answer is infinity. Nothing can change that answer.</p>
<p>erm, the answer is undefined, not infinity</p>
<p>Well, it's undefined, but in the more specific cases of infinity or negative infinity, even though still technically not defined, it is often preferable to make it clear that it's infinity.
At least, that's what I've learned.</p>
<p>Yes I know it is undefined, but as ^ said, it is often easier, for calculations, etc, to take it as infinity. But I should have made myself clearer. sorry</p>
<p>haha, you calc nerds. xD</p>