by-considering-different-paths-of-approach-show-that-the-functions-in-exercises-41-48-have-no-limit-

Question

By considering different paths of approach, show that the functions in Exercises $41-48$ have no limit as $(x, y) \rightarrow(0,0)$ .
$$f(x, y)=-\frac{x}{\sqrt{x^{2}+y^{2}}}$$

Stephen Hobbs · Accepted Answer

with multi variable limits instead of approaching just from the left and the right of one value, as we did with single variables. We&#x27;re going approach from many different directions of lines. So the one line that we could check is the line why you call zero or the approaching from the X axis? We could say so the limit as y approaches zero, would end up getting us, uh, negative X over square root of X squared would be X and then simplifying this, we&#x27;d end up getting negative one. Now, if we can get another value, we can prove that this limit does not exist. So, for instance, let&#x27;s approach from the line. Why equals X or another way to say that is as X approaches. Why that would get us negative y on top, divided by the square root. Oh, uh, why squared plus y squared. Which one end up giving us two y squared? Taking the square root, we&#x27;d end up getting negative. Why over square root to why? And then simplifying these, we&#x27;d end up getting a fraction of negative one over route to And because we&#x27;re getting two different values approaching from two different lines towards this, this 20.0 we know that this limit does not exist

Stephen Hobbs · Answer

approaching this limit from two different lines we could first look at approaching from the line Y equals zero. We&#x27;re taking the limit as why approaches zero, and doing so will knock out the wise where a term in the denominator leaving us with X to the fourth, divided by X to the fourth, which would simplify to just one. Now, if we can come up with another line of approach that does not equal this, we know that the limit does not exist. And so if we can get anything other than one, we know that that&#x27;s going to prove this. So, for instance, let&#x27;s look at this in terms of a common technique. Is trying to combine all of your like terms so that you get your variables to cancel so we could do that if we let Why approach X squared, for instance, and the reason this is helpful is because the numerator is X to the fourth, and then the denominator is X to the fourth plus substituting X squared for why it be another term of extra fourth and combining like terms that would get us to X to the fourth and the denominator X to the fourth and the numerator. And then, of course, Thes two would end up canceling. And so we really end up with a limit of 1/2 because that&#x27;s different than the limit that we got from a different approach. This tells us that this limit at 00 does not exist. It&#x27;s like getting two different values from the left and right handed limit for a single variable problem.

Stephen Hobbs · Answer

analyzing this limit from two different lines of approach. We could first look at the line of approach from the Y axis, where X is approaching zero and doing so that would knock out the X terms. And we&#x27;d be left with negative Y squared, divided by Y squared and simplifying. This would get us negative one. If we can, um, evaluate from a different line of approach and get a different value, we know that this limit does not exist and we could do so I actually approaching from the X axis this time or letting why the limit is why approaches zero. And in this case, the white terms get knocked out and we have X to the fourth divided by X to the fourth, which, of course, just simplifies so one. And since we get to different values, this is analogous to getting to different values with a single variable from the left and right hand approach. And so we know that this limit does not exits

Stephen Hobbs · Answer

analyzing this limit from two different approaches will start by evaluating the limit as X approaches zero or the line X equal zero, in other words. And then we&#x27;ll end up getting zero divided by wise, where and in this case we would end up simplifying that 20 considering another approach. If we tried something a little bit different. The limit as why approaches X squared, for instance, then substituting here will end up getting X squared times x word divided by X to the fourth waas Uh, next. And whenever we can get like terms, then this is a great way to have a secondary, you know, approach to a limit that is otherwise may be difficult, simplify or or find another path. And so, in this case, again, we&#x27;ll end up with, um X to the fourth, divided by two X to the fourth. And now, because of these life terms, they&#x27;ll go ahead, cancel. And so the second limit would equal 1/2. Because we have two different values here. We know that this limit does not exist

Stephen Hobbs · Answer

we could analyse this problem from two different lines. First off, we could analyze this from the Line X equals zero or as a limited as X approaches zero and substituting that value in we done up getting why, over why and that would simplify toe one. Now it doesn&#x27;t work so well if we substitute y equals zero, because we&#x27;ll get undefined by dividing by zero. But we could also try something where we try to get like terms, for instance. So let&#x27;s say we take the limit from the line. Why approaches X squared that would end up getting us keeping the X weird plus substitute X word for why all divided by why? Which again it&#x27;s X squared. And now we can combine like terms in the numerator. And so we&#x27;d end up getting two X squared, divided by X squared and needs to would simplify to value of to. And since we&#x27;re getting two different values coming from two different approaches, we know that this limit does not exist

Stephen Hobbs · Answer

analyzing this limit from two different approaches. We could analyse the limit along the line y equals X or, in other words, the limit is X approaches. Why and doing so, replacing X with why we get y squared up top divided by absolute value of y squared in the denominator. And then, of course, the removing the absolute value. We know that this would just get us a value of one looking at a different line of approach. Let&#x27;s suppose this time that we looked at the line y equals negative X or X equals negative why? And this would end up getting us negative. Why squared up top, replacing that X with a negative why and the absolute value of negative Why? Where in the bottom, of course, the absolute value would remove the negative in the denominator. So this would get us negative y squared over positive y squared, which would simplify to negative one because we get to different values from two different lines of approach. This is equivalent to this function, not existing or, um, this limit, not existing. It&#x27;s analogous to the one sighted limits from the left and the right not being an agreement for a single variable like we saw in calculus before

Problem

By considering different paths of approach, show …

Problem 41 Medium Difficulty

By considering different paths of approach, show that the functions in Exercises $41-48$ have no limit as $(x, y) \rightarrow(0,0)$ .
$$f(x, y)=-\frac{x}{\sqrt{x^{2}+y^{2}}}$$

Answer

$\lim_{y\to 0} f(x,y) = -1$, but, $\lim_{x\to y} f(x,y) = \frac{-1}{\sqrt{2}}$

Related Courses

Thomas Calculus

Related Topics

Discussion

Top Calculus 3 Educators

Lily A.

Heather Z.

Kayleah T.

Michael J.

Calculus 3 Courses

Vectors Intro

Vector Basics - Example 1

Recommended Videos

Watch More Solved Questions in Chapter 14

Video Transcript

George B. Thomas, Jr. Maurice D. Weir, Joel Hass

Thomas Calculus

Lily A.

Heather Z.

Kayleah T.

Michael J.

Vectors Intro

Vector Basics - Example 1

What do you need help with?

Textbooks

Ask a question

Courses

AI Tutor

$\lim_{y\to 0} f(x,y) = -1$, but, $\lim_{x\to y} f(x,y) = \frac{-1}{\sqrt{2}}$

Thomas Calculus

Lily A.

Heather Z.

Kayleah T.

Michael J.

Vectors Intro

Vector Basics - Example 1

George B. Thomas, Jr. Maurice D. Weir, Joel HassThomas Calculus

Lily A.

Heather Z.

Kayleah T.

Michael J.

George B. Thomas, Jr. Maurice D. Weir, Joel Hass

Thomas Calculus