# Evaluating limit for this function

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1. Sep 28, 2016

### toforfiltum

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. 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.

Last edited: Sep 28, 2016
2. Sep 28, 2016

### Staff: Mentor

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.

3. Sep 28, 2016

### toforfiltum

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

Thanks!

4. Sep 28, 2016

### Staff: Mentor

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 continous 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.

5. Sep 28, 2016

### toforfiltum

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

Thanks!

6. Sep 28, 2016

### SammyS

Staff Emeritus
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 !)

7. Sep 28, 2016

### toforfiltum

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