Multivariable Limits assignment

In summary, the conversation is about finding limits for various functions and the difficulties the person is facing in finding the correct answer. They mention that for the first two limits, they have tried different paths but still cannot find the correct answer. They also discuss the concept of limits and how finding the limit involves evaluating the function at different points. Finally, they mention that in order to get credit for the problem, all the answers need to be correct.
  • #1
I can't get this problem right and it's part of a web-based assignment that I have to submit. In order to get credit for the problem, all of my answers need to be right; I've tried many different times and I can't seem to figure out what I am doing wrong. Here it is with my explanations as to why I think I'm right; please, can someone find my mistake?

Find the following limits. If a limit diverges to infinity, state your answer as "INF" (without the quotation marks). If it diverges to negative infinity, state your answer as "MINF".If it diverges without being infinity or negative infinity, state your answer as "DIV".

lim(x,y)->(0,0) (3x^2 + y^2)/(x^4 + 5 y^4)
Limit does not exist because if I try to go through different paths, I get different limits.

lim(x,y)->(0,3) ysin(x)/x
= 3
by separating the two functions y & sin(x)/x and evaluating them seperately

lim(x,y)->(5,0) (3x + ln(1+xy) ) / (1+x+y)
just by plugging in the values (function is continuous everywhere)

lim(x,y)->(0,0) (x^2 * y^2) / (x^4 + 5y^4)
Limit does not exist once again because of different paths I took = different limits

Thanks a lot =)
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  • #2
You're confusing how to find different paths. Instead, what you're doing is finding the limits at different points. Think of the limit as this:

[tex]\lim_{\textbf{r}\rightarrow (0,0)}f(x,y)[/tex].

Here, [tex]\textbf{r}[/tex] just represents some arbitrary path along the surface to that point.
  • #3
I understand what you mean but, for example, if I take the fourth limit:
lim(x,y)->(0,0) (x^2 * y^2) / (x^4 + 5y^4)

If I evaluate this when x=y then I end up with (x^4)/(6x^4) which gives a limit of 1/6.
However, if I evaluate this at (x,0) which should also work, I get a limit of 0/(x^4) = 0
Because both these limits (along different paths) are different, then the entire limit can not possibly converge.
  • #4
from what I can see, you're correct. You sure you typed it in right?
  • #5
yeah, I am sure.. but to get it right they all have to be right.. I am pretty confident about the last two limits.. its mainly the first two that are bugging me
  • #6
Along which paths did you take the limits to evaluate the first one?

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