Calculate the indicated limits
- Thread starter Bulma
- Start date
D
Deleted member 4993
Guest
Hello everyone
I was wondering if I did this correctly, thanks in advance!
<link removed>
I copied the OP's work from the image site.
Last edited by a moderator:
pka
Elite Member
- Joined
- Jan 29, 2005
- Messages
- 11,971
No you did not.
The first rule of doing limits is:
Never, Never, Never, Never use a substitution of the limiting variable value.
Do you understand that \(\displaystyle \displaystyle{\large\lim _{x \to 2}}~\frac{{x - 2}}{{2 - x}} = - 1~???\)
If you substitute the \(\displaystyle x=2\) what do you get? Is it defined? WHY?
What if you substitute \(\displaystyle x=2.001\) then \(\displaystyle 1.999~?\).
Are the two results almost the same value? What is that value?
IS THAT VALUE THE LIMIT?
- Joined
- Feb 4, 2004
- Messages
- 16,582
Eventually, the image will be taken down. To preserve the meaning in this (eventually archived) thread, here's what that image contains:
6. Calculate the indicated limits: (a) \(\displaystyle \displaystyle \lim_{x\, \rightarrow\, -3}\, \)\(\displaystyle \dfrac{x^2\, +\, 3x}{x^2\, -\, 9}\,\) (b) \(\displaystyle \displaystyle \lim_{x\, \rightarrow\, 2}\, \)\(\displaystyle \dfrac{4\, -\, x^2}{x^2\, -\, x\, -\, 2}\)
My working:
(a) \(\displaystyle \displaystyle \lim_{x\, \rightarrow\, -3}\, \)\(\displaystyle \dfrac{x^2\, +\, 3x}{x^2\, -\, 9}\)
\(\displaystyle \dfrac{(-3)^2\, +\, 3(-3)}{(-3)^2\, -\, 9}\, =\, \dfrac{9\, -\, 9}{9\, -\, 9}\, =\, \dfrac{0}{0}\)
\(\displaystyle \dfrac{x\, (x\, +\, 3)}{(x\, -\, 3)\,(x\, +\, 3)}\, =\)
\(\displaystyle \displaystyle \lim_{x\, \rightarrow\, -3}\, \)\(\displaystyle \dfrac{-3}{(-3)\, -\, 3}\, =\, \dfrac{-3}{-6}\)
\(\displaystyle \displaystyle \lim_{x\, \rightarrow\, -3}\, \)\(\displaystyle =\, \dfrac{1}{2}\)
(b) \(\displaystyle \displaystyle \lim_{x\, \rightarrow\, 2}\, \)\(\displaystyle \dfrac{4\, -\, x^2}{x^2\, -\, x\, -\, 2}\)
\(\displaystyle \dfrac{4\, -\, (2)^2}{(2)^2\, -\, (2)\, -\, 2}\, =\, \dfrac{4\, -\, 4}{4\, -\, 2\, -\, 2}\, =\, \dfrac{0}{0}\)
\(\displaystyle \displaystyle \lim_{x\, \rightarrow\, 2}\, \)\(\displaystyle \dfrac{4\, -\, x^2}{x^2\, -\, x\, -\, 2}\)
\(\displaystyle \dfrac{(2\, +\, x)\, (2\, -\, x)}{(x\, +\, 1)\, (x\, -\, 2)}\)
\(\displaystyle \displaystyle \lim_{x\, \rightarrow\, 2}\, \)\(\displaystyle \dfrac{(2\, +\, 2)\, (2\, -\, 2)}{(2\, +\, 1)\, (2\, -\, 2)}\, =\, \dfrac{4}{3}\)
\(\displaystyle \displaystyle \lim_{x\, \rightarrow\, 2}\,\)\(\displaystyle =\, \dfrac{4}{3}\)
- Joined
- Feb 4, 2004
- Messages
- 16,582
No, for many reasons.
This last line above confirms that you must take the limit, because you cannot evaluate.
This is the factorization you did back in algebra when you were working with rational functions, doing the graphing "with the hole", where you had one x-value for which the original function was not defined. But this factorization is in no way "equal" to the limit that follows. Many graders would stop grading at this "equals", with everything following being ignored as incorrect.
No. Since the expression is not, as you showed, actually defined at x = -3, you cannot take the limit by evaluating at x = -3; the function is not defined at x = -3. Instead, you have to use valid mathematical logic, such as "anything divided by itself, as long as it isn't zero, has '1' as its limiting value". This is how limits get around the "not defined at x = (whatever)" problem.
Your book (and your class notes) will contain at least one worked example of this sort. Review it for pointers:
. . . . .\(\displaystyle \begin{align} \displaystyle \lim_{x\, \rightarrow\, -3}\, \dfrac{x^2\, +\, 3x}{x^2\, -\, 9}\, &=\, \lim_{x\, \rightarrow\, -3}\, \dfrac{x\, (x\, +\, 3)}{(x\, -\, 3)\, (x\, +\, 3)}\,
\\ \\ &=\, \lim_{x\, \rightarrow\, -3}\, \left(\dfrac{x}{x\, -\, 3}\right)\, \left(\dfrac{x\, +\, 3}{x\, +\, 3}\right)\end{align}\)
...and so forth.
i think the op is actually going on a good direction by performing direct substitution first, and when he got 0/0 he did factoring (evident on first question) and tried to cancel a common factor but the problem is that in the 2nd question, he did a small mistake since anything multiplied to zero is zero, meaning what he did is wrong. The correct direction is to do direct substition then try factoring the numerator and denominator, if that fails then try rationalizing (hint; in rationalizing when evaluating limits, you can use either the numerator or the denominator)
forgot to add that in that particular question (#2) just factor out -1 in the denominator. check what you get
forgot to add that in that particular question (#2) just factor out -1 in the denominator. check what you get
Last edited:
i forgot to add...
if you tried all of the above, but got c/0; c= any real number not equal to zero. then you can use a theorem where;
if c>1 and the lim of the denominator approaches zero from positive values then you get lim of f is positive infinity
if c<1 and the lim of the denominator approaches zero from positive values then you get lim of f is negative infinity
if c>1 and the lim of the denominator approaches zero from negative values then you get lim of f is negative infinity
if c<1 and the lim of the denominator approaches zero from negative values then you get lim of f is positive infinity
if you tried all of the above, but got c/0; c= any real number not equal to zero. then you can use a theorem where;
if c>1 and the lim of the denominator approaches zero from positive values then you get lim of f is positive infinity
if c<1 and the lim of the denominator approaches zero from positive values then you get lim of f is negative infinity
if c>1 and the lim of the denominator approaches zero from negative values then you get lim of f is negative infinity
if c<1 and the lim of the denominator approaches zero from negative values then you get lim of f is positive infinity