Limits of Complex Functions: Final Exam Practice

[itsagoal89]

compute the limit lim as x,y --> (0,0) of (x+y)*ln(x2+y2)

i would use polar coordinates if (x+y) was (x^2+y^2) but that not being the case is messing with me.

and than another one is the same parameters, but of (x^2*y)/(x^4+y^2). for this one i attempted to do the partial deriv of both x and y, but than i didn't what to do when i arrived at each component. that then made me beg the question "was i even allowed to do that"

Someone help, and set me straight, these are final exam practice problems.

 

[LCKurtz]

compute the limit lim as x,y --> (0,0) of (x+y)*ln(x2+y2)

i would use polar coordinates if (x+y) was (x^2+y^2) but that not being the case is messing with me.

Polar is still a good idea. You can overestimate (x+y) = r(cos(θ)+sin(θ))...

 

[LCKurtz]

compute the limit lim as x,y --> (0,0)...

and than another one is the same parameters, but of (x^2*y)/(x^4+y^2). for this one i attempted to do the partial deriv of both x and y, but than i didn't what to do when i arrived at each component. that then made me beg the question "was i even allowed to do that"

Someone help, and set me straight, these are final exam practice problems.

Perhaps this one was find the limit if it exists. See if you can find a couple paths giving different limits.

 

[itsagoal89]

thanks kurt, but i don't know how i would, what would (x,y)->(0,0) change to? and i get stuck here...

lim (r(cos(a)+sin(a))*ln(r^2)
xy->(0,0)

 

[LCKurtz]

thanks kurt, but i don't know how i would, what would (x,y)->(0,0) change to? and i get stuck here...

lim (r(cos(a)+sin(a))*ln(r^2)
xy->(0,0)

Of course, you want absolute values in there. But you can bound the trig functions easily enough, and (x,y) → (0,0) is the same as r → 0, so it is just a problem about rLn(r²) in 1-D. Try L'Hospital on it after getting it in the right form.

 

[cereeal]

Or use squeeze method:
|xln(x^2+y^2)| $$\leq$$ f(x,y) $$\leq$$ 2|xln(x)|

 

[itsagoal89]

i hate the squeeze theorem. i am never going to use it again after calc3, i had a teacher ruin it for me,


KURT-
r -> 0 good call, i figured that part on my owner about five minutes after. but i don't know if its just my awful teacher, but i am not seeing what you are saying in "you can bound it easy enough". without a denominator how do i use l'hopital because he is the man for identifying that rule and i would love to use it. show me what form i would be taking the limit of.

 

[LCKurtz]

thanks kurt, but i don't know how i would, what would (x,y)->(0,0) change to? and i get stuck here...

lim (r(cos(a)+sin(a))*ln(r^2)
xy->(0,0)

Of course, you want absolute values in there. But you can bound the trig functions easily enough, and (x,y) → (0,0) is the same as r → 0, so it is just a problem about rLn(r²) in 1-D. Try L'Hospital on it after getting it in the right form.

i hate the squeeze theorem. i am never going to use it again after calc3, i had a teacher ruin it for me,


KURT-
r -> 0 good call, i figured that part on my owner about five minutes after. but i don't know if its just my awful teacher, but i am not seeing what you are saying in "you can bound it easy enough". without a denominator how do i use l'hopital because he is the man for identifying that rule and i would love to use it. show me what form i would be taking the limit of.

Bound the trig functions each by 1 and write the r times the ln(r) like this:

$$r\ln(r) = \frac {\ln(r)}{\frac 1 r}$$

This gives you an ∞/∞ form you can use L'H on.

 

[itsagoal89]

kurt, rln(r^2) is a dream to solve. but converting the "r*(cos+sin)" to just the r is where you are losing me. i don't understand what you mean by "bound each by 1"
this is what i am picturing when you say that
-1=<cos=<1
-1=<sin=<1

yet i don't know how ^that helps at all.
wr
or do you mean that in order to be allowed to write it as "1"*"R"*"LN(R^2)", that i first have to say "if cos + sin =<1, than the equation is "", because the sum of the squares of each are one."?

 

[itsagoal89]

i also tried not converting it, and got a beast of a problem when i used l'hospital

 

[LCKurtz]

$$|r(\cos\theta + \sin\theta)\ln(r^2)|\le |r|(|\cos\theta|+|\sin\theta|)|2\ln(r)| \le 4|r\ln(r)| = 4\left|\frac {\ln(r)}{\frac 1 r}\right|\rightarrow 0$$