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1. In my textbook (Stewart's Calculus), the video tutor solutions for some problems use the squeeze theorem to determine the limit of a function. For example: Find lim (x, y) → (0, 0) x2y3 2x2 + y2. The typical solution I keep seeing involves taking the absolute value of f(x, y) and then using some properties of inequalities to deduce the ...
limr→0 r2cos4 θsin4 θ (cos2 θ +r2sin4 θ)3 = 0 cos6 θ lim r → 0 r 2 cos 4 θ sin 4 θ (cos 2 θ + r 2 sin 4 θ) 3 = 0 cos 6 θ. Now the limit mentioned above would be equal to zero if and only if the denominator is different than zero. Hence we need to calculate the limit in the case where θ = π2 + kπ θ = π 2 + k π and compare it ...
lim (x, y) → (0, 0) f(x, y) = lim r → 0 + 4r2log(r) and the limit on the right is entirely a single variable calculus limit, to which we can apply all our single variable limit theorems including l'Hôpital's rule (which will be needed in this case). This method may seem to have few applications, however in combination with the squeeze ...
For each ϵ> 0, let δ ≤ min (ϵ 6, (ϵ 6)1 4). Then, starting with | 5r3cos3(θ) − r4cos2(θ)sin2(θ) | and working through the inequalities as above, we come to the expression 5r3 + r4. If ϵ ≥ 6, then ϵ 6 ≥ (ϵ 6)1 4 and therefore r <(ϵ 6)1 4. Thus, 5r3 + r4 <5(ϵ 6)3 4 + ϵ 6. Since ϵ 6 ≥ 1, we have (ϵ 6)3 4 ≤ ϵ 6, so 5(ϵ ...
This is usually the first resort, and if the paths are chosen judiciously, you will obtain two different answers, which implies the nonexistence of the limit, because for the limit to exist, it must have the same value along every possible path. Note that this test can only be used to show nonexistence: to prove a limit exists requires more work.
Finding examples of two different approaches giving different limits (in the case that the limit doesn't exist) is usually easier in the original $(x,y)$ coordinates. The point of polar coordinates (as I see it) is to have a tool for proving that the limit is what you think it is (in the case when the limit exists). $\endgroup$ –
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Currently I have been learning the limits of two-variable functions. I know that in order to show the non-existence of a given limit, we need to select two distinct paths for testing. If the two outcomes are different, the limit does not exist. Yet, I don't know the exact way for path selection. To be more specific, let's refer to the example ...
8. The definition of the limit is as follows: lim (x, y) → (x0, y0) f(x, y) = L if and only if for all ϵ> 0 there exists a δ> 0 such that. | f(x, y) − L | <ϵ. whenever 0 <√(x − x0)2 + (y − y0)2 <δ. NOTE: Alternatively, lim (x, y) → (x0, y0) f(x, y) = L if and only if for all ϵ> 0 there exists a δ> 0 such that. | f(x, y) − L ...
This is usually the first resort, and if the paths are chosen judiciously, you will obtain two different answers, which implies the nonexistence of the limit, because for the limit to exist, it must have the same value along every possible path. Note that this test can only be used to show nonexistence: to prove a limit exists requires more work.
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related to: multivariable limit calculator- 3495 U.S. 1, Princeton, New Jersey · Directions · +1 609-520-8385
