Checking Answer: Evaluate the following Limits

S

SeekerOfDragons

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Oct 8, 2009
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Trying to verify that my answers are correct to the following problems:

Limit as X --> 3 (X^2 - 3X) / (X^2 - 9)

Limit as X --> 3 [X(X - 3) ] / [(X - 3)(X + 3)
Limit as X --> 3 X / (X + 3)

3 / (3 + 3)
3/6

**** 1/2

Limit as X --> Infinity (X^2 + X + 2) / (4X^3 - 1)

(X^2 + X + 2) * [ (1/X^3) / (1/X^3) ] / (4X^3 - 1) * [ (1/X^3) / (1/X^3) ]
(X^2/X^3 + X/X^3 + 2/X^3) / (4X^3/X^3 - 1/X^3)
(1/X + 1/X^2 + 2/X^3) / (4 - 1/X^3)
(0 + 0 + 0) / (4 - 0)
0 / 4

**** 0

For this also, looking at the powers can tell you what it evaluates to:
if Powers are =, then evaluates to value of coefficients of x variable.
if power of Numerator > power of Denominator, evaluates to Infinity
if power of Denominator > Power of Numerator, evaluates to 0

Just wanted to verify that the process I used in showing my work is accurate and an acceptable method of solving the problem.

Another side note/question. I may be wrong since it's been a little bit, but is it possible to solve the above by finding the Derivative of the equations? I've tried to find the Derivative on both by using the Quotient rule and by converting and using the Product rule, but somewhere along the way, my math isn't working out right. any assistance in that regard for my own benefit would be appreciated as well...

thanks,

r/

SoD
 
\(\displaystyle Are \ you \ familiar \ with \ the \ Marqui \ Guillaume \ Francois \ Antione \ De \ L'Hopital?\)

\(\displaystyle Both \ of \ the \ above \ limits \ can \ be \ solved \ using \ his \ method.\)
 
Big Glenn,

I had forgotten about L'Hopital. Looking over my notes, I rediscovered it and can see how it would be much easier than how I was doing it in many cases.

Using that method on Limit as X --> 3 (X^2 - 3X) / (X^2 - 9) I got:

Original limit = 0/0 so can use L'Hopital's Rule so:
(2X - 3) / 2X
(2(3) - 3) / 2(3)
(6 - 3) / 6
3 / 6

**** 1/2 Agrees with my original, but can be much easier on more complex problems

On the second problem Limit as X --> Infinity (X^2 + X + 2) / (4X^3 - 1)

Original Limit comes to Infinity/Infinity so can use L'Hopital's rule so:
(2X + 1) / 12X^2

Still evaluates to Infinity / Infinity so apply rule again:
2 / 24X

**** No longer evaluates to 0/0 or Infinity / Infinity so can no longer apply L'Hopital's rule. Evaluating the final equation 2 / 24X has 2 being divided by a bigger and bigger number (infinity) such that it is approaching 0 which again agrees with my original answer, but is much, much easier to come up with.

Is my understanding of the concept and answers regarding these two problems correct? I'm assuming my answers are correct since I came to the same conclusion using both methods...

Really appreciating your help over the last few days confirming my understanding and answers.

r/

SoD
 
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