Use polar coordinates to find the limit. [If (r, θ) are polar coordinates of the point (x, y) with r ≥ 0, note that r → 0+ as (x, y) → (0, 0).] (If an answer does not exist, enter DNE.)
lim (x, y)→(0, 0) [(3e^−x^2− y^2) − 3]/
(x^2 + y^2)
in polar coordinates, we have
lim (r->0) (3e^-(r^2)-3)/r^2
= lim -6re^(-r^2))/2r
= lim -6e^(-r^2)/2
as r-> 0, lim = -3
Well, well, well, we have a polar limit on our hands! Let's dive right in and see what we can find.
First, let's express those Cartesian coordinates in terms of polar coordinates. Since (x, y) → (0, 0), this means that r → 0+ as well. So we can write x = rcosθ and y = rsinθ.
Now let's substitute these expressions into the given function. We get:
lim (r, θ)→(0, 0) [(3e^−(rcosθ)^2− (rsinθ)^2) − 3]/((rcosθ)^2 + (rsinθ)^2)
Simplifying a bit, we have:
lim (r, θ)→(0, 0) [3e^−r^2(cos^2θ + sin^2θ) − 3]/(r^2(cos^2θ + sin^2θ))
Good news, everyone! The term (cos^2θ + sin^2θ) is just equal to 1, so we can simplify further to:
lim (r, θ)→(0, 0) [3e^−r^2 − 3]/r^2
Now let's analyze this bad boy. As r approaches 0, we see that the numerator (3e^−r^2 − 3) approaches 3e^0 − 3 = 3 - 3 = 0.
But the denominator r^2 approaches 0 as well. And we all know that dividing by zero is a big no-no! So unfortunately, the limit does not exist (DNE).
I'm sorry if that's not the punchline you were hoping for, but sometimes limits can be a bit...limiting!
To find the limit, we can convert the expression to polar coordinates.
Let's replace x and y with their polar coordinate counterparts rcos(θ) and rsin(θ), respectively.
The expression becomes:
lim (r, θ) → (0, 0) [(3e^−(rcos(θ))^2−(rsin(θ))^2) − 3]/(r^2(cos^2(θ) + sin^2(θ)))
Simplifying this expression:
lim (r, θ) → (0, 0) [(3e^−r^2(cos^2(θ) + sin^2(θ))) − 3]/(r^2)
Since e^-r^2 approaches 1 as r approaches 0, we have:
lim (r, θ) → (0, 0) [(3-3)]/(r^2)
Simplifying further:
lim (r, θ) → (0, 0) 0/(r^2) = 0
Therefore, the limit of the given expression as (x, y) approaches (0, 0) is 0.
To find the limit using polar coordinates, we need to express the given expression in terms of polar coordinates (r, θ).
Let's start by substituting x = rcosθ and y = rsinθ into the expression:
lim (r, θ)→(0, 0) [(3e^−(rcosθ)^2− (rsinθ)^2) − 3]/(r^2cos^2θ + r^2sin^2θ)
Simplifying the expression:
lim (r, θ)→(0, 0) [(3e^−(r^2cos^2θ+ r^2sin^2θ)) − 3]/(r^2(cos^2θ + sin^2θ))
Since cos^2θ + sin^2θ = 1, we can simplify further:
lim (r, θ)→(0, 0) [(3e^−r^2) − 3]/(r^2)
Now, we can evaluate the limit as r approaches 0+:
lim r→0+ [(3e^−r^2) − 3]/(r^2)
To find this limit, we can apply L'Hôpital's Rule, which states that if we have a limit of the form 0/0, we can take the derivative of the numerator and denominator and then evaluate the limit again. Let's differentiate the numerator and denominator with respect to r:
Numerator: d/dx (3e^−r^2) = -6re^−r^2
Denominator: d/dx (r^2) = 2r
Taking the limit again:
lim r→0+ [(-6re^−r^2)/(2r)]
Simplifying further:
lim r→0+ [-3e^−r^2]
As r approaches 0, the exponential term e^−r^2 approaches 1, so the limit becomes:
-3(1) = -3
Therefore, the limit as (x, y) approaches (0, 0) of the given expression is -3.