Evaluating limit for this function

[toforfiltum]

Homework Statement

Function is $lim_{(x,y,z) \rightarrow (0,\sqrt\pi,1)} \ e^{xz} \cos y^2 - x$

Homework Equations

The Attempt at a Solution

As $x \rightarrow 0$ along $y= \sqrt \pi, z=1$, $f(x,y,z)= -1$

As $y \rightarrow 0$ along $x=0, z=1$, $f(x,y,z) = -1$

As $z \rightarrow 1$ along $x=0, y= \sqrt \pi$, $f(x,y,z) = -1$

From what I have done, the limit looks like it is -1. I'm wondering if there are any other functions I could use to show that it is true. Or this is not needed at all, because of the argument that the function is continuous at the point $(0,\sqrt \pi, 1)$? If so, the above steps are not needed at all, right?

Thanks.

 

[Mark44]

Homework Statement

Function is $lim_{(x,y,z) \rightarrow (0,0,0)} \ e^{xz} \cos y^2 - x$

Homework Equations

The Attempt at a Solution

As $x \rightarrow 0$ along $y= \sqrt \pi, z=1$, $f(x,y)= -1$

As $y \rightarrow 0$ along $x=0, z=1$, $f(x,y) = -1$

As $z \rightarrow 1$ along $x=0, y= \sqrt \pi$, $f(x,y) = -1$

From what I have done, the limit looks like it is -1. I'm wondering if there are any other functions I could use to show that it is true. Or this is not needed at all, because of the argument that the function is continuous at the point $(0,\sqrt \pi, 1)$? If so, the above steps are not needed at all, right?

The function in the limit is continuous at (0, 0, 0), so there's not need to be concerned about different paths to this point. Also, why are you concerned about the point $(0, \sqrt{\pi}, 1)$? It's nowhere near to the point at which the limit is being taken.

 

[toforfiltum]

The function in the limit is continuous at (0, 0, 0), so there's not need to be concerned about different paths to this point. Also, why are you concerned about the point $(0, \sqrt{\pi}, 1)$? It's nowhere near to the point at which the limit is being taken.

Oops, sorry! My typo. The question did indeed ask me to evaluate the limit at $(0, \sqrt\pi, 1)$.

Thanks!

 

[Mark44]

Oops, sorry! My typo. The question did indeed ask me to evaluate the limit at $(0, \sqrt\pi, 1)$.

OK, then that makes more sense, but it doesn't change what I said. The exponential part ($e^{xz}$) is continuous for every real value of x and z; the cosine function is continuous for all real y (and y²) and the simple polynomial x is continuous for all real x. Sums, differences, and products of these functions are also continuous for all real x, y, and z.

Where you run into problems is with quotients or other operations that produce undefined results.

 

[toforfiltum]

OK, then that makes more sense, but it doesn't change what I said. The exponential part ($e^{xz}$) is continuous for every real value of x and z; the cosine function is continuous for all real y (and y²) and the simple polynomial x is continuous for all real x. Sums and differences of these functions are also continuous for all real x, y, and z.

So in this case, the limit at the point is equal to the inputs of the function.

Thanks!

 

[SammyS]

Homework Statement

Function is $lim_{(x,y,z) \rightarrow (0,\sqrt\pi,1)} \ e^{xz} \cos y^2 - x$

Homework Equations

The Attempt at a Solution

As $x \rightarrow 0$ along $y= \sqrt \pi, z=1$, $f(x,y)= -1$

As $y \rightarrow 0$ along $x=0, z=1$, $f(x,y) = -1$

As $z \rightarrow 1$ along $x=0, y= \sqrt \pi$, $f(x,y) = -1$

From what I have done, the limit looks like it is -1. I'm wondering if there are any other functions I could use to show that it is true. Or this is not needed at all, because of the argument that the function is continuous at the point $(0,\sqrt \pi, 1)$? If so, the above steps are not needed at all, right?

Thanks.

It looks like there are some typos in your post.

From what's in the attempt, it appears that you actually are interested in $\displaystyle\ lim _{(x,\,y,\,z)\to (0,\, \sqrt \pi ,\, 1)}\ f(x,y,z)\,,\$ where $\displaystyle\ f(x,y,z)= e^{xz} \cos (y^2) - x \ .$

Where you have $\ f(x,y)\,,\$ I think you mean to have $\ f(x,y,z)\ .$(It took me wayyyyyy tooooo long to type this in !)

 

[toforfiltum]

It looks like there are some typos in your post.

From what's in the attempt, it appears that you actually are interested in $\displaystyle\ lim _{(x,\,y,\,z)\to (0,\, \sqrt \pi ,\, 1)}\ f(x,y,z)\,,\$ where $\displaystyle\ f(x,y,z)= e^{xz} \cos (y^2) - x \ .$

Where you have $\ f(x,y)\,,\$ I think you mean to have $\ f(x,y,z)\ .$(It took me wayyyyyy tooooo long to type this in !)

Oh gosh, yes. I meant $f(x,y,z)$. So sorry for the carelessness. I guess my brain is too tired of multivariable calculus.