Find the limit, if it exists. (If an answer does not exist, enter DNE.)
lim (x, y)→(0, 0)
of (x^2 + y^2)/[((x^2 + y^2 +64)^.5)-8]
one good way to tackle xy limits is to convert to polar coordinates, then let r->0
f(r,theta) = r^2/(sqrt(r^2+64)-8)
now, using lHospital's Rule,
lim = 2r/[2r/(2sqrt(r^2+64))]
= 2sqrt(r^2+64))
= 2*8 = 16
Since the function does not involve theta, the limit is the same from all directions.
but there is no lHospitals rule in multivariable
Let's find the limit of the given function as (x, y) approaches (0, 0).
First, we notice that the denominator involves the square root of (x^2 + y^2 + 64), which means that the distance from the point (x, y) to the origin is always greater than or equal to 8.
To find the limit, we need to consider the limit along different paths approaching (0, 0).
Approaching along the x-axis (y = 0):
lim (x, 0)→(0, 0) [(x^2 + 0^2)/[((x^2 + 0^2 + 64)^.5)-8]]
= lim (x, 0)→(0, 0) (x^2/[sqrt(x^2 + 64) - 8])
If we substitute y = 0 into the function, we get (x^2/[sqrt(x^2 + 64) - 8]). As x approaches 0 along the x-axis, the expression in the numerator becomes 0, and the expression in the denominator becomes -8. Therefore, the limit is 0/(-8) = 0.
Approaching along the y-axis (x = 0):
lim (0, y)→(0, 0) [(0^2 + y^2)/[((0^2 + y^2 + 64)^.5)-8]]
= lim (0, y)→(0, 0) (y^2/[sqrt(y^2 + 64) - 8])
Similarly, if we substitute x = 0 into the function, we get (y^2/[sqrt(y^2 + 64) - 8]). As y approaches 0 along the y-axis, both the numerator and denominator approach 0. Applying L'Hopital's rule, we differentiate the numerator and the denominator with respect to y:
lim (0, y)→(0, 0) [2y/[2sqrt(y^2 + 64)]]
= lim (0, y)→(0, 0) [y/sqrt(y^2 + 64)]
Now, we can substitute y = 0 into the expression, and we get 0/sqrt(0^2 + 64) = 0/8 = 0.
Since the limit along the x-axis and the y-axis both approach 0, the limit as (x, y) approaches (0, 0) exists and is equal to 0.
To find the limit of the given function as (x, y) approaches (0, 0), we can use the polar coordinate representation of points in the xy-plane. In polar coordinates, x = r cosθ and y = r sinθ, where r represents the distance from the origin and θ represents the angle from the positive x-axis.
Let's substitute these expressions into the function:
lim(r, θ)→(0, 0) [(r cosθ)^2 + (r sinθ)^2] / [((r cosθ)^2 + (r sinθ)^2 + 64)^.5 - 8]
Simplifying this expression, we get:
lim(r, θ)→(0, 0) (r^2 cos^2θ + r^2 sin^2θ) / [((r^2 cos^2θ + r^2 sin^2θ + 64)^.5) - 8]
Notice that the numerator r^2 is common to all terms. We can factor it out:
lim(r, θ)→(0, 0) r^2 * (cos^2θ + sin^2θ) / [((r^2 cos^2θ + r^2 sin^2θ + 64)^.5) - 8]
Using the identity cos^2θ + sin^2θ = 1, we further simplify:
lim(r, θ)→(0, 0) r^2 / [((r^2 + 64)^.5) - 8]
Now, let's examine the expression inside the square root: (r^2 + 64). As r approaches 0, the value of r^2 will approach 0. So, we have:
lim(r, θ)→(0, 0) r^2 / [(64)^.5 - 8]
Simplifying further, we get:
lim(r, θ)→(0, 0) r^2 / (8 - 8)
Notice that the denominator is 0, leading to an undefined value.
Therefore, the limit does not exist.