# Thread: Help with limits. I can do the limit, but I can't explain "which theorem i used"

1. ## Help with limits. I can do the limit, but I can't explain "which theorem i used"

1. lim as x goes to 2...(x^2 - 4)/(x-2)

2. lim as x goes to 0...sin 2x + x^2 cos 5x

3. lim as x goes to 0 from the negative side... [(square root)x^2]/x

as ive said, ive gotten all the limits, but I need to tell which theorem i used.... we can only use like the 5 most basic ones, and i cant see how any of them apply to any of these

2. Identities like these...

$\displaystyle \frac{x}{x} = 1$

$\displaystyle \sqrt{x^2}=x$

... are not consequences of some particular theorem but only a question of elementary 'good sense'...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

3. thats exactly what i thought, but they explicitly say to note what limit theorems you are using

the ones we got are:
basically the 4 operations using limits f and g...eg lim f(x) + g(x)= lim f(x) + lim g(X)

squeeze principal

continuity of indefinite inegrals

4. Originally Posted by twostep08
thats exactly what i thought, but they explicitly say to note what limit theorems you are using

the ones we got are:
basically the 4 operations using limits f and g...eg lim f(x) + g(x)= lim f(x) + lim g(X)

squeeze principal

continuity of indefinite inegrals

(x^2 - 4)/(x-2) = (x-2)(x+2)/(x-2) . Now, from the very definition of limit, when x --> 2 we explicitly rule out the value x = 2, so that above we can divide by x-2 and get very simply lim (x + 2) = 4 when x --> 2

For the enxt one you use arithmetic of limits: both sin(2x) and x^2*cos 5x have limit zero when x --> 0: the first one directly by continuity of sin x. and the second one because x^2 --> 0 clearly and cos 5x is bounded.

The last one is nice: since we're working with negative x's, Sqrt(x^2) = |x| ==> the limit wanted is lim |x|/x = lim (-1) = -1 when x --> 0-

Tonio

5. Originally Posted by chisigma
Identities like these...

$\displaystyle \frac{x}{x} = 1$

$\displaystyle \sqrt{x^2}=x$

... are not consequences of some particular theorem but only a question of elementary 'good sense'...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
Neither of those are true.
$\displaystyle \frac{x}{x}= 1$ only for $\displaystyle x\ne 0$ and $\displaystyle \sqrt{x^2}= |x|$, not x.

6. Originally Posted by twostep08
1. lim as x goes to 2...(x^2 - 4)/(x-2)
Theorem: if f(x)= g(x) for all x except x= a, then $\displaystyle \lim_{x\rightarrow a} f(x)= \lim_{x\rightarrow a} g(x)$.

2. lim as x goes to 0...sin 2x + x^2 cos 5x
Theorem: if f is continuous at x= a, then $\displaystyle lim_{x\rightarrow a}f(x)= f(a)$
Well, more the definition of "continuous" than a theorem. And sine and cosine are continuous for all x.

3. lim as x goes to 0 from the negative side... [(square root)x^2]/x
$\displaystyle \sqrt{x^2}= |x|$ which is -x for x negative. That is, $\displaystyle \frac{\sqrt{x^2}}{x}= \frac{-x}{x}$ which is -1 as long as x is not 0. Now use the theorem I cited for (1).

as ive said, ive gotten all the limits, but I need to tell which theorem i used.... we can only use like the 5 most basic ones, and i cant see how any of them apply to any of these

7. Originally Posted by HallsofIvy
$\displaystyle \sqrt{x^2}= |x|$ which is -x for x negative.
This cannot be true, can it? Aren't you are saying $\displaystyle |x|=-x$ when x is negative? This flies in the face of absolute values!

8. No it's true like it is. If $\displaystyle x<0$ then $\displaystyle x^2>0$ and so does the absolute value...

9. Originally Posted by pflo
This cannot be true, can it? Aren't you are saying $\displaystyle |x|=-x$ when x is negative? This flies in the face of absolute values!

Not at all: this is the exact definition of abs. value for negative values...pay attention: NEGATIVE values , so -x is.....

Tonio

10. Originally Posted by HallsofIvy
Neither of those are true.
$\displaystyle \frac{x}{x}= 1$ only for $\displaystyle x\ne 0$ and $\displaystyle \sqrt{x^2}= |x|$, not x.
I strongly agree on the principle that everibody is free to express his own opinion, no matter which opinion is...

... but I also strongly prefer avoid to wasting time in tedious discussions because...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

11. Originally Posted by chisigma
I strongly agree on the principle that everibody is free to express his own opinion, no matter which opinion is...

... but I also strongly prefer avoid to wasting time in tedious discussions because...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
The post by HoI (which you quoted) is not an opinion, it is mathematical fact. Details like this are important. It is never a waste of time to note and correct errors of understanding. These sorts of errors are extremely common - it is not a waste of time to make sure that the OP has the correct mathematical understanding. Far from being tedious, the discussion in this thread is extremely valuable.

Edit: Thread closed. The discussion was getting off-point. As far as I can see, the questions have been satisfactorily answered. twostep08, if you require more help please pm me.