Calculus 3 Question.Can someone help me with this question? I have a problem figuring it out.

Compute the limit if they exist

1. As (x,y) approaches (0,0) of (e^xy - 1)/y

2. As (x,y) approaches (0,0) of (cos (xy) - 1) / (x^2 y^2)

3. As (x,y) approaches (0,0) of (x^3 - y^3) / (x^2 + y^2)

4. As (x,y) approaches (0,0) of (sinxy) / y

1) lim((x,y)→(0,0)) (e^(xy) - 1)/y
= lim((x,y)→(0,0)) [(1 + (xy) + (xy)^2/2! + ...) - 1] / y, using the series for exp
= lim((x,y)→(0,0)) [(xy) + x^2 y^2/2! + ...] / y
= lim((x,y)→(0,0)) (x + x^2 y/2! + ...)
= 0.
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2) lim((x,y)→(0,0)) (cos (xy) - 1) / (x^2 y^2)
= lim((x,y)→(0,0)) [(1 - (xy)^2/2! + (xy)^4/4! - ...) - 1] / (x^2 y^2), using the series for cos
= lim((x,y)→(0,0)) [-(xy)^2/2! + (xy)^4/4! - ...] / (xy)^2
= lim((x,y)→(0,0)) [-1/2! + (xy)^2/4! - ...]
= -1/2.
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3) Use polar coordiantes:
lim((x,y)→(0,0)) (x^3 - y^3) / (x^2 + y^2)
= lim(r→0+) r^3 (cos^3(t) - sin^3(t)) / r^2
= lim(r→0+) r (cos^3(t) - sin^3(t))
= 0, by the Squeeze Theorem.

Details:
-1 - 1 ≤ cos^3(t) - sin^3(t) ≤ 1 + 1 for all t
==> -2r ≤ r(cos^3(t) - sin^3(t)) ≤ 2r for all t.

Since lim(r→0+) ±2r = 0, the result now follows.
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4) lim((x,y)→(0,0)) sin(xy)/y
= lim((x,y)→(0,0)) x sin(xy)/(xy)
= [lim((x,y)→(0,0)) x] * [lim((x,y)→(0,0)) sin(xy)/(xy)]
= 0 * 1, since xy→0 as (x,y)→(0,0)
= 0.

I hope this helps1

1. Suppose (x,y) -> (0,0) along the line y = x. Then the function is [(e^(x^2) - 1]/x; this has the form 0/0 as x -> 0, so we may use L'Hospital's Rule. The quotient of derivatives is 2xe^(x^2)/1, and the limit is 0.

3. Suppose (x,y) -> (0,0) along x = y. The function looks like 0/2x^2, and you can again use L'H to find the limit is 0.

Get the idea?