Crap, what an embarrassing mistake to make... Regardless, the picture is still largely meaningless. Does 81/3 = âˆš81 x 3 really merit the title of " ******* math"?
Actually the two sided limit is undefined. But as x approaches 8 from the left, the limit is -infinity. As x approaches 8 from the right, the limit goes to +infinity.
Sources: I'm Chinese. I'm born with this knowledge imbued in me.
The question is asking what the equation approaches as x gets closer to the limit but not what it is AT the limit. so in both it would be +- infinity because you are dividing over a number that is infinitely small in + or - depending on what direction you approach from. [url deleted]
is a graph of both of these questions (not really but it shows the idea)
No. Math has notation just like English has grammar. The limit doesnt exist because coming from the right and coming from the left the limits are different, meaning the overall limit does not exist.
Pic related. It's the same way with math. If it was limx->8- [1/(x-8)] then it would be true
This is why schools should just use standard limit notation. (parentheses) for approaches without bound, and [brackets] for reaches. so this would be (-âˆž,5) (5,âˆž) because the function never has an X value of negitive infinity, 5, or positive infinity.
Actually that's only true if you're assuming x=8, so y(x-8) would be infinity x zero which equals 1. What's more, in the case that (x-8) is zero or even less than one, y would have to be greater than one for the whole thing to equal 1.
I remember seeing the picture as a kid. At that time I was looking at it for 30 min trying to figure out how it is done. In the end I have come to a conclusion that I just can t do it.
On this very day in exactly 1 h I have a partial exam in limes and demand and supply functions. Is it irony that I see this picture today ?
This is actually a common misconception. You see, because of the way the lim matrix is constructed, the vector between the unknown value of x and the constant 5 is unparallel to the equation's infinitly changing result. Thus the final sum must be inverded from the one currently shown.
In this case, I direct your eyes to the attached .png file to my left for the actual result.
In other words, the limit doesn't exist in R, which is the space that is generally assumed if none is given. The equation could still hold if the limit were taken in a different space, but one would generally expect something like that to be specified.
I think you're confusing upper and lower bounds/continuity with limit theory... the function is discontinuous and unbounded in the neighbourhood of x = 8 but the limit can still be equated to infinity... this just proves discontinuity.
Wrong.
Limit x->8+ (from the right) 1/(x-8) = infinity.
Limit x->8- (from the left) 1/(x-8) = -infinity
If limit x->a+ =/= limit x->a- then limit x->a Does not exist.
in the first one it is asking for a limit (lim) or the value of y as x approaches 8. The close you get to 8 (8.0001 for instance), the smaller the number you are dividing by, which inscreases the number you are dividing. so when you get really really small like .00000000000001 it approaches infinity.
for the second one it is asking about the value of y as the line of the second equation's graph approaches 5 on the x axis. The close you get to 5, the smaller the denominator becomes and the lager the numerator becomes after division; the limit is infinity again.
the joke is based on the two upright 8's and the "sideways 8" (infinity sign) in the first one and there being 5 upright 5's in the second so someone put a sideways 5 as the answer
Nice explanation... Though im kinda confused. At my university we use a different way to write that, as here is no information about the direction. You cant see here whether its going from negativ raising up to 8, or from positiv down to 8. So it could be plus or minus infinity.
Sorry for the bad english, I'm majoring in maths, not english^^
That is correct. As there is no indication of direction it is asking for the overall limit, which does not exist. It does not exist because the limit from the left and the limit from the right are different. In order for the statement to be correct it should specify that they are taking the limit from the right.
While the concept that as you divide by a number approaching zero your product will be closer to infinity makessense, it just doesn't work that way, unfortunately.
This means that X/0 = infinity
So therefore {1,2,3,4,....} over zero = infinity aswell
which, of course is the problem, as 1 is now = 2 = 3 = 4 = 5 = 6 etc
1/x-8 in particular is not going to be a very happy graph when x = 8. Why? Because you're dividing by zero. That's what the limit is. It's asking what the nature of the graph will be when it approaches a situation where the function will divide by zero.
Further investigation would do a left and right side limit. 1/a very small positive and 1 over a very small negative. You know your fractions right. Anything divided by something less than zero is going to become bigger, and these are infinitely small numbers... so 1 becomes infinite. The only difference is that depending on the sign of the small number it will either approach positive or negative infinite...
the limit of the exuation 1/(x-8) as x approaches the value 8 is infinity. Reason being is that as x approaches 8 the equation becomes closer 1/0. 1/0 is infinity kinda. Same goes for 5.