Question

Ok, I don't know if I should put this in the calc1 subforums but LaTex is below

Answer 1

\({\lim_{x \rightarrow 0}}\frac{1-\cos{h}}{h}\) at \(\large h=\pm0.1,\pm0.01,\pm0.001\)

Answer 2

this is basically asking \[\lim_{x \rightarrow 0^+}\frac{ 1-\cos(h) }{ h }\] just plugin the points and test them.

Answer 3

^Yep, since the function is defined at all those values, just plug them in and evaluate.

They want you to be able to see that the original limit approaches zero.

Answer 4

Oh, this managed to get bumped? lol it didn't show up on mine --

Sorry, I was saving this for another problem

Answer 5

Subforums?
Do people still use those? :3

Answer 6

Okay so
\[\lim_{x\rightarrow-3^{-}}\frac{2e^{x}}{(x+3)^{3}}\]

they want the infinite limit which I calculated via graph to be \(-\infty\)

but the answer that the professor gave was "as \(x\rightarrow3^{-}...\)"
\(2e^{x}\rightarrow2e^{-3}\) (positive #s)
\((x+3)^{3}\rightarrow0^{-}\) (negative #s)

Answer 7

the limit is \(x\rightarrow-3^{-}\) btw

Answer 8

Use \Large to make it more legible\[\Large \lim_{x\rightarrow-3^{-}}\frac{2e^{x}}{(x+3)^{3}}\]

Answer 9

oh yeah sorry;;

so idk, did I get it at least half right with \(f(x)\rightarrow-\infty\)?

Answer 10

Yes, it's negative infinity, since the numerator is positive and denominator is negative and approaching zero.

Answer 11

I honestly thought it was more of an end behavior question idk. But then there's this --

and the answer to another question \(\Large{\lim_{x\rightarrow\frac{\pi}{2}^{+}}\sec{x}}\)

Answer 12

End behaviour means when x is approaching pos. or neg. infinity.

Answer 13

So end behavior \(\ne\) infinite limits?

Answer 14

http://www.wolframalpha.com/input/?i=limit+(2e%5Ex)%2F%5B(x%2B3)%5E3%5D+as+x-%3E-3-

Answer 15

End behaviour refers to when x is approaching pos. or neg. infinity. Nothing to do with a limit that is equal to a value approaching infinity.

Answer 16

... okay, then what does the WA link mean by \(-\infty\)?

Answer 17

It's all so confusing sorry ;;

Answer 18

That the function value approaches neg. infinity, when x approaches -3 from left.

Answer 19

So end behavior is the x-value and infinite limit is the y-value or output?

Answer 20

*x-value/input

Answer 21

End behaviour will look like\[\Large \lim_{x \rightarrow \infty}\]or\[\Large \lim_{x \rightarrow -\infty} \]

Answer 22

And yes.

Answer 23

OH.
Well that was simpler than I thought... wow

Answer 24

I'm still not sure why the problem with e has a two-part answer though

Answer 25

\(\color{blue}{\text{Originally Posted by}}\) kittiwitti1
\[\lim_{x\rightarrow-3^{-}}\frac{2e^{x}}{(x+3)^{3}}\]
they want the infinite limit which I calculated via graph to be \(-\infty\)

but the answer that the professor gave was "as \(x\rightarrow3^{-}...\)"
\(2e^{x}\rightarrow2e^{-3}\) (positive #s)
\((x+3)^{3}\rightarrow0^{-}\) (negative #s)
\(\color{blue}{\text{End of Quote}}\)

Answer 26

He's just describing the result.

Answer 27

Oh, so it's still \(f(x)\rightarrow-\infty\) then?

Answer 28

\[\Large \lim_{x\rightarrow-3^{-}}\frac{2e^{x}}{(x+3)^{3}} = -\infty\]and then what he wrote is just describing why

Answer 29

http://prntscr.com/cd78wi

Oh, alright. Thank you for clarifying

Answer 30

If anyone sees this, btw...
how do I calculate asymptotes with a function like \(\Large{\frac{2e^{x}}{(x+3)^{3}}}\)?

Answer 31

In the land of math, dividing by 0 is a crime.\[\large\rm \frac{1}{0}\qquad \leftarrow\text{this is not a number}\]
Asymptotic behavior leads to this though.
\[\large\rm \frac{1}{x}\qquad\text{asymptote at x=0}\]
\[\large\rm \frac{1}{x-2}\qquad \text{asymptote at x=2}\]

Answer 32

To state simply, the denominator tells us about asymptotes.
Whatever value causes the denominator to be 0 leads to an asymptote.

Answer 33

So in your problem, \(\large\rm (x+3)^3=0\)
Solving for x will tell us that special number that causes the denominator to be 0.
Or in this case, simply eye-balling it, because -3 and 3 give us 0, ya? :)

Answer 34

@zepdrix I was actually wondering about the numerator in terms of finding any possible horizontal or slant asymptotes, but that was also helpful :p

Answer 35

I don't think you can get slant/oblique asymptotes from non-polynomial stuff. Those types of asymptotes show up when the degree of the numerator is exactly one larger than the degree of the denominator.

Some examples of graphs with oblique asymptotes:\[\large\rm \frac{x^4+4}{x^3+7},\qquad\qquad \frac{x^2}{x+1}\]

So you don't need to worry about that when exponential stuff is thrown into the mix.

Answer 36

Horizontal asymptotes show up at the far ends of the graph,
end behavior as you were talking about before :)

Answer 37

Soooo ... when the numerator has a wacky \(x\) then disregard all HA/SA rules? hahah

Answer 38

I mean... I WA'd it and got this
http://www.wolframalpha.com/input/?i=(2e%5Ex)%2F%5B(x%2B3)%5E3%5D+asymptotes

Answer 39

Disregard SA rules, yes.
But take for example something with wacky x stuff like this,\[\large\rm \frac{e^{-7x}}{x^2+8}\]
As x approaches the end of the graph (on the right side),
e^{-7x} is approaching zero much much faster than x^2+8 is blowing up.
So the numerator is winning toward 0, so horizontal asymptote at y=0.

I'm not sure what techniques you've covered...
It would some fancier tools than algebra to determine horizontal asymptotes for exponential stuff.

Answer 40

I guess I didn't need to come up with another example.
Yours would have worked just fine.

It's a comparison, a race, between the numerator and denominator.
You want to know who is winning the race.

Answer 41

When you plug x=-3bajillion into the numerator and denominator separately,
\(\large\rm 2e^{-3bajillion}=\tiny \text{really really small}\)
\(\large\rm (-3bajillion+3)^3=\text{ehh sort of large}\)

Numerator is winning to zero.

Answer 42

Ok that's bad explanation maybe XD haha.

Answer 43

\(\Large{bajillion}\) LOL

Answer 44

Ok to summarize:

Vertical Asymptotes: Set denominator equal to 0.
Horizontal Asymptotes: Look at the ends of the graph.
Slant Asymptotes: None if wacky stuff included.

Answer 45

(mom's insulting me with rather crude things and telling me this site is "the reason you get C D F" ok)

Can you calculate HA from the fcn alone?

Answer 46

lolol

Answer 47

fcn?

Answer 48

Function. Abbreviated lol

Answer 49

You can get a clear idea of what the HA should be by plugging in a really really big x value (positive or negative).

But you'll have to use this idea of limits to be completely satisfied.

Answer 50

; - ;

Answer 51

I just used the normal n < m rule for HA = 0 in this function though\[\lim_{x\rightarrow8^{+}}\frac{1}{x-8}\]

Answer 52

Ok fair enough :)
You could also plug in a really big x value to get an idea of what's going on off in the distance.

If we plug x=7batrillion... or no no, let's use a more sensible number like x=10007. So we get,\[\large\rm \frac{1}{x-8}\quad=\quad \frac{1}{10007-8}=\frac{1}{9999}\approx.0001\]It's getting really close to 0 when x is big.

Answer 53

Hm, makes sense

Answer 54

meow -_-

Answer 55

Okay so... the n < m cannot apply to the function that has numerator with exponential \(x\)?

Answer 56

Correct.\[\large\rm \frac{x^n+3}{x^m+94}\]Where n<m, collapses to 0 for big x.


\[\large\rm \frac{(x^n+3)e^{5x}\cos(7x)}{x^m+94}\]But this is a whole something else going on here D:

Answer 57

what the heck is that
a failed experiment x'D

Answer 58

ya haha :D

Answer 59

ok well I'm glad I didn't have that lol

Answer 60

Cool beanios.