Question

I was sick when we went over limits in class, and so I'm a bit behind. How do i find the limit as x approaches 0 from the negative side of 1/x- 1/sqrroot v(x)

Answer 1

\[\lim_{x \rightarrow 0^-}= \left( \frac{ 1 }{ x }-\frac{ 1 }{ \sqrt{x} } \right)\]

Answer 2

Would I just use the calculator for this? because my teacher hates it when we use the calculator, so I watned to know if it's possible algebraically.

Answer 3

yes it's possible :) one sec.

Answer 4

Nice, thanks.

Answer 5

Approaching zero from the left?
Hmm this function isn't defined on the left side of zero.
Did you post it correctly? D:

See how there is a square root?
We can't plug a negative into a root.

Answer 6

That's how it is on the paper, It's weird.

Answer 7

It's probably a typo, so I'll just ask the teacher tomorrow. So instead could we try a different problem? I just want to get some examples of what to do so i can try it on my own.

Answer 8

Yah let's try approaching from the right side of zero and see what happens.\[\Large\bf \lim_{x \rightarrow 0^+}\quad \frac{ 1 }{ x }-\frac{ 1 }{ \sqrt{x} }\]

Answer 9

So I can't plug in 0 or it's undefined, so I'll have to try something else right?

Answer 10

Yah I guess we can't plug 0 directly in since it would cause a problem.
Umm this method might be a little sloppy, but at least it'll give us an idea.

When we get closer to zero, which term is going to be larger, 1/x or 1/sqrtx ?

Like compare, \(\Large\dfrac{1}{.05}\) to\(\Large\dfrac{1}{\sqrt{.05}}\)

Answer 11

Grr It's been so long since I've tried to do these algebraically lol :)
I guess I shouldn't shortcut like that :p

Answer 12

It's alright. Short cuts are great.

Answer 13

\[\frac{ 1 }{ .05 } is bigger\]

Answer 14

Good, so that tells us that \(\Large \dfrac{1}{\sqrt x}\) is approaching zero faster than \(\Large\dfrac{1}{x}\),

Both terms are gonna explode, blow up to infinity, but the sqrt is getting there slower.
Which means we'll have a positive quantity when we subtract.

Answer 15

Not the most elegant explanation -_-
I think there's a way to do this more algebraically, it's just slipped my mind at the moment.

Answer 16

So I'm trying to say that our limit is giving us something like this,\[\Large \infty_{bigger}-\infty\quad=\quad +\infty\]
That's sloppy bad notation, don't ever write it like that lolol.
We don't ever say that something "equals" infinity. :p

Answer 17

You don't have any other examples we can try? :D

Answer 18

i do actually. How about?
\[\lim_{x \rightarrow (\frac{ \pi }{ 2 })^-} \frac{ \tan x +3}{ \tan^2x }\]

Answer 19

Hmm these are tricky :p
So when x is pi/2, tangent gives us the indeterminate form 1/0.

If our angle x is slightly smaller than pi/2 (that's what it means to approach from the left), then tangent is giving us a positive value as it approaches the form 1/0.
Hmm I can't remember how to deal with these, I might have to call the big dog over to help us :P

The indeterminate form is telling us that tangent is approaching \(\Large \infty\).
So if we just compare the numerator and denominator,
The denominator is getting closer to \(\Large\infty\) much faster due to the square, yes? D:

Answer 20

@Zarkon @jim_thompson5910 @mathstudent55

Answer 21

\[\Large \lim_{x \rightarrow (\frac{ \pi }{ 2 })^-} \frac{ \tan x +3}{ \tan^2x }\]

Answer 22

Alright, so that makes sense that the denominator is getting bigger faster. I've got to eat some dinner now, so I'll be back in a little bit.

Answer 23

k

Answer 24

I'm back.

Answer 25

These problems are a bit trickier when the top AND bottom are approaching infinity.
So we want to compare the top and bottom and see which one is "winning".

In this case our denominator is winning (getting there faster),
So we can say that our expression is approaching something like this:\[\Large \frac{C}{\infty}\]The numerator just isn't getting there fast enough, so it's value will be insignificant compared to the denominator.

Answer 26

okay. And since the denominator gets so close to infinite that means the whole thing is getting smaller and smaller right?

Answer 27

Yessss, good good.
So it looks like our expression is approaching zero overall.

NORMALLY we would want to be careful and keep track of our `signs`.
But in this case, -0 = +0, so the signs won't matter.

Answer 28

cool. Thanks.