Question

Find the limit, if it exists. lim_{(x, y) ? (0, 0)} rac{xy}{sqrt{x^2 + y^2}} Step 1 To examine lim_{(x, y) ? (0, 0)} rac{xy}{sqrt{x^2 + y^2}}, first approach (0, 0) along the x-axis. On this path, all points have y = 0. Step 2 Since y = 0 for all points on this path, then we have lim_{(x, y) ? (0, 0)} rac{xy}{sqrt{x^2 + y^2}} = lim_{x ? 0} rac{x ? 0}{sqrt{x^2 + 0}} = 0. Step 3 Similarly, approaching along the y-axis yields a limit equal to 0. Since these two limits are the same, we will examine another approach path. Approach (0, 0) along the curve y = x^2. When x is positive, we have lim_{(x, y) ? (0, 0)} rac{xy}{sqrt{x^2 + y^2}} = lim_{x ? 0} rac{x ? x^2}{sqrt{x^2 + x^4}} = lim_{x ? 0} rac{x^2}{sqrt{1 + x^2}}. When x is negative, we have lim_{(x, y) ? (0, 0)} rac{xy}{sqrt{x^2 + y^2}} = lim_{x ? 0} rac{x ? x^2}{sqrt{x^2 + x^4}} = lim_{x ? 0} -rac{x^2}{sqrt{1 + x^2}}.

          Find the limit, if it exists.
lim_{(x, y) ? (0, 0)} rac{xy}{sqrt{x^2 + y^2}}
Step 1
To examine lim_{(x, y) ? (0, 0)} rac{xy}{sqrt{x^2 + y^2}}, first approach (0, 0) along the x-axis.
On this path, all points have y = 0.
Step 2
Since y = 0 for all points on this path, then we have
lim_{(x, y) ? (0, 0)} rac{xy}{sqrt{x^2 + y^2}} = lim_{x ? 0} rac{x ? 0}{sqrt{x^2 + 0}} = 0.
Step 3
Similarly, approaching along the y-axis yields a limit equal to 0. Since these two limits are the same, we will examine another approach path. Approach (0, 0) along the curve y = x^2.
When x is positive, we have
lim_{(x, y) ? (0, 0)} rac{xy}{sqrt{x^2 + y^2}} = lim_{x ? 0} rac{x ? x^2}{sqrt{x^2 + x^4}} = lim_{x ? 0} rac{x^2}{sqrt{1 + x^2}}.
When x is negative, we have
lim_{(x, y) ? (0, 0)} rac{xy}{sqrt{x^2 + y^2}} = lim_{x ? 0} rac{x ? x^2}{sqrt{x^2 + x^4}} = lim_{x ? 0} -rac{x^2}{sqrt{1 + x^2}}.

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Find the limit, if it exists.
lim(x, y) ? (0, 0) racxysqrtx^2 + y^2
Step 1
To examine lim(x, y) ? (0, 0) racxysqrtx^2 + y^2, first approach (0, 0) along the x-axis.
On this path, all points have y = 0.
Step 2
Since y = 0 for all points on this path, then we have
lim(x, y) ? (0, 0) racxysqrtx^2 + y^2 = limx ? 0 racx ? 0sqrtx^2 + 0 = 0.
Step 3
Similarly, approaching along the y-axis yields a limit equal to 0. Since these two limits are the same, we will examine another approach path. Approach (0, 0) along the curve y = x^2.
When x is positive, we have
lim(x, y) ? (0, 0) racxysqrtx^2 + y^2 = limx ? 0 racx ? x^2sqrtx^2 + x^4 = limx ? 0 racx^2sqrt1 + x^2.
When x is negative, we have
lim(x, y) ? (0, 0) racxysqrtx^2 + y^2 = limx ? 0 racx ? x^2sqrtx^2 + x^4 = limx ? 0 -racx^2sqrt1 + x^2.

Added by Daniel L.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Darshan Maheshwari

12/19/2021

Step 1

On this path, all points have \(y = 0\). Show more…

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Find the limit, if it exists. lim_{(x, y) → (0, 0)} rac{xy}{sqrt{x^2 + y^2}} Step 1 To examine lim_{(x, y) → (0, 0)} rac{xy}{sqrt{x^2 + y^2}}, first approach (0, 0) along the x-axis. On this path, all points have y = 0. Step 2 Since y = 0 for all points on this path, then we have lim_{(x, y) → (0, 0)} rac{xy}{sqrt{x^2 + y^2}} = lim_{x → 0} rac{x ∙ 0}{sqrt{x^2 + 0}} = 0. Step 3 Similarly, approaching along the y-axis yields a limit equal to 0. Since these two limits are the same, we will examine another approach path. Approach (0, 0) along the curve y = x^2. When x is positive, we have lim_{(x, y) → (0, 0)} rac{xy}{sqrt{x^2 + y^2}} = lim_{x → 0} rac{x ∙ x^2}{sqrt{x^2 + x^4}} = lim_{x → 0} rac{x^2}{sqrt{1 + x^2}}. When x is negative, we have lim_{(x, y) → (0, 0)} rac{xy}{sqrt{x^2 + y^2}} = lim_{x → 0} rac{x ∙ x^2}{sqrt{x^2 + x^4}} = lim_{x → 0} -rac{x^2}{sqrt{1 + x^2}}.

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Transcript

00:01 And this question, we have to find a limit if it exists and that's multivariable.

00:07 So this is what we have to pay attention on.

00:11 If you're approaching from the y -axis, that he is a limit equal to zero, and since two limits are same, we'll examine another approach, and now we are approaching along y -equal to x -square.

00:20 So it's as simple as replacing y by x -square.

00:24 So if we replace y -by -x -square, that's going to look like limit x -y, extend to 0 0.

00:32 And if x, y is replaced with x squared, then x remains as it is and this y is rewritten with x squared.

00:39 And then the denominator that will be x square plus y is again written as x square and whole square remains as it is.

00:46 So the numerator will become a limit.

00:51 This will become x cube over root of x square plus x raise to four uh which is nothing but limit x cube over and in the denominator we can simplify this by taking x square out so we have x square times one plus x square root that's what we can take in fact okay i'm not sure if we have to simplify the denominator or not but i think we can okay, all right.

01:25 Now, if we see carefully, this is where we have x squared.

01:29 It means they are canceling one of the x cube.

01:31 So if we divide, if we want to cancel one of the x cube, what we have to do is one among the x cube, one of x in x cube.

01:40 We have to take x square out from a denominator to write it as x square times one plus x square.

01:45 And now if we take this x square out, this x square out, so that's going to be one of the x is going to come out...

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