Question

"Only one of these calculations are correct. Which one? Why are the others wrong? Explain your answers." (Calculations below in a minute).

Answer 1

\[\lim_{x \rightarrow 0}x \ln(x) = 0 * (-\infty) = 0 \]\[\lim_{x \rightarrow 0}x \ln(x) = 0 * (-\infty) = -\infty \]\[\lim_{x \rightarrow 0}x \ln(x) = \lim_{x \rightarrow 0} \frac{ \ln (x) }{ (1/x) } = \frac{ -\infty }{ \infty } = -1\]\[\lim_{x \rightarrow 0} x \ln (x) = \lim_{x \rightarrow 0} \frac{ \ln (x) }{ (1/x) } = \lim_{x \rightarrow 0} \frac{ (1/x) }{ (-1/x ^{2} )} = \lim_{x \rightarrow 0} (-x) = 0 \]

I don't really understand at all how the natural log works in this context; i've heard that natural logs of zero can behave as infinities. For the last two, I just have no idea what they did, unless they just took the reciprocal of one of them to make it possible to apply L'Hopital's rule. I would apply the product rule and come up with this.

Answer 2

(And I feel overwhelmingly like the last one is correct, but not sure how I would explain the infinities with the natural logs, if that is even remotely correct.)

Answer 3

\[0 * (-\infty) = 0\] is false , log or no log

Answer 4

this is a limit statement
and you cannot assert that \[0 * (-\infty) = 0\] any more than you can say
\[0 * (-\infty) =-\infty\]

Answer 5

this statement
\[\frac{ -\infty }{ \infty } = -1\] is also false

Answer 6

so really the false ones have nothing at all to do with the log

Answer 7

(Give me just a minute if i'm not responding immediately, by the way, thank you very much already)

Answer 8

Why is the first one false? I thought (Again, infinities are weird so that's where i'm guessing it stems from) zero times anything is zero.

Answer 9

here is a simple example
\[\lim_{x\to 0^+}\frac{1}{x^2}=\infty\] and
\[\lim _{x\to 0}x^4=0\] what can we conclude about
\[\lim_{x\to 0}\frac{1}{x^2}\times x^4\]?
what happens if you replace \(x^4\) by just \(x\)? all those limit statements are wrong

Answer 10

a limit statement is just that, a statement about limits, not about numbers
while it is true that \(0\times a=0\) if \(a\) is a number, it is not true that \(0\times \infty=0\) because this is not a statement about numbers, it is statement about what happens as \(x\) approaches some number (or maybe as \(x\to \infty\) )

Answer 11

i am sure you do not think
\[\frac{\infty}{\infty}=1\]
think about
\[\lim_{x\to \infty}\frac{x^2+2}{x^3}\]
that limit is clearly 0, even though the limit in the numerator and the denominator are both \(\infty\)

Answer 12

Is the bottom degree larger? I think that's a three, but I can't tell.

Answer 13

and similarly
\[\lim_{x\to \infty}\frac{2x-3}{5x}=\frac{2}{5}\]

Answer 14

yeah, but the simple point is you cannot reason out what a limit is by says "zero times anything is zero" or "infinity divided by infinity is one, because any number divided by itself is 1" or any other such nonsense

Answer 15

so at the risk of repeating myself, the false statements above have nothing to do with the log
they are false on their face

Answer 16

I guess? I mean, you're telling me that it's nonsense, and you're correct, but i'm not quite clear on what the other logical path is, then. And well, no, I don't know what infinity/infinity would be. What would you reason that to be.

Answer 17

you have to check to see what the limit is, just like you check that
\[\lim_{x\to \infty}\frac{2x+1}{5x}=\frac{2}{5}\]
you cannot assume it is anything until you compute it

Answer 18

sometimes it is hard, sometimes you are just supposed to know them, like
\[\lim_{x\to 0}\frac{\sin(x)}{x}=1\]

Answer 19

in this case it looks like you are supposed to find it using l'hopital

Answer 20

Yeah, i'm aware of that. But how do you reason with infinities? You could apply L'Hopital's Rule or other stuff for the example you gave, but how do you deal with infinities in the absence of variables?

Answer 21

(Just on a different note. I don't know much about infinities and it would do me good to understand them better.)

Answer 22

Like how would you evaluate \[\frac{ \infty }{ \infty } \] Ignoring limits, shouldn't that be 1?

Answer 23

i am not sure what the question is about "absence of variables" but that is precisely why there are rules like l'hopital, to allow you to compute these things

Answer 24

something like
\[0\times \infty\] or
\[\frac{\infty}{\infty}\] are called "indeterminate forms" precisely because you do not know what they are

Answer 25

but they are just shorthand in any case
there is no such number as \[0\times \infty\] what it is shorthand for is if
\[\lim_{x\to a}f(x)=0\] and
\[\lim_{x\to a}g(x)=\infty\] then we say the FORM of
\[\lim_{x\to a}f(x)g(x)=0\times \infty\] it doesn't say what the limit actually is

Answer 26

Okay. I wasn't aware \[0*\infty \]Was an indeterminate form until now, I was aware of \[\frac{ 0 }{ 0 }\]and\\[\frac{ \infty }{ \infty } \]until now. I think i'm starting to get it; I knew which one was correct, just had trouble clearly outlining why the others were wrong. Thanks, anyways.

Answer 27

yw