Finding Limits of functions of Several Variables

 For each of the following functions find the limit (at the given point), if it exists, or show that the limit does not exist. Solutions at the end.

> f1 := (x,y) -> 8*x^2*y^2/(x^4+y^4); x1:=0; y1:=0;

```
2  2
x  y
f1 := (x,y) -> 8 -------
4    4
x  + y

x1 := 0

y1 := 0
```

> f2 := (x,y) -> sin(x+y)/(x+y); x2:=0; y2:=0;
```
sin(x + y)
f2 := (x,y) -> ----------
x + y

x2 := 0

y2 := 0
```

> f3 := (x,y) -> (x^2+y^2-2*x-2*y)/(x^2+y^2-2*x+2*y+2);x3:=1;y3:=-1;
```
2    2
x  + y  - 2 x - 2 y
f3 := (x,y) -> -----------------------
2    2
x  + y  - 2 x + 2 y + 2

x3 := 1

y3 := -1
```

> f4 := (x,y,z) -> x*exp(z) + ln(2*x-y); x4:=2;y4:=3;z4:=0;
```
f4 := (x,y,z) -> x exp(z) + ln(2 x - y)

x4 := 2

y4 := 3

z4 := 0
```

> f5 := (x,y) -> (x^3+x*y^2)/(x^2+y^2);x5:=0;y5:=0;
```
3      2
x  + x y
f5 := (x,y) -> ---------
2    2
x  + y

x5 := 0

y5 := 0
```

Solutions

For each case we check first if the limits along all straight lines are the same if they are we check for continuity. If the function is not continuous then we check other paths (parabolas and cubics) if we still don't get different limits we go to the epsilon, delta definition.

Example 1

> Limit(f1(x, y1 + m*(x-x1)),x=x1) = limit(f1(x, y1 + m*(x-x1)),x=x1);

```
4  2          2
x  m          m
Limit  8 ---------- = 8 ------
x -> 0    4    4  4          4
x  + m  x      1 + m
```

since the limit depends on the slope of the line then there is no limit.

Example 2

> Limit(f2(x, y2 + m*(x-x2)),x=x2) = limit(f2(x, y2 + m*(x-x2)),x=x2);

```
sin(x + m x)
Limit  ------------ = 1
x -> 0    x + m x
```

 all the straight lines give the same limit 1. Notice that f2 is the composition of G with F where,

> F := (x,y) -> x+y; G := t -> sin(t)/t; (G@F)(x,y);

```
F := (x,y) -> x + y

sin(t)
G := t -> ------
t

sin(x + y)
----------
x + y
```

and since
The limit of a composition is the composition of the limits
when the limits exist of course. In our case the limits of the two functions F and G clearly exist so the limit of the composition also exits and it is 1.

Example 3

> Limit(f3(x, y3 + m*(x-x3)),x=x3) = limit(f3(x, y3 + m*(x-x3)),x=x3);

```
2                    2
x  + (- 1 + m (x - 1))  - 2 x + 2 - 2 m (x - 1)      infinity
Limit  ----------------------------------------------- = --------------
x -> 1    2                    2                                     2
x  + (- 1 + m (x - 1))  - 2 x + 2 m (x - 1)     signum(1 + m )
```

 so all the limits along straight lines give infinity. Don't get confused with the denominator, it is in fact 1. Maple is careful that "m" may be a complex number with negative square so that the sign may change. To show that the limit exists in this case, we could use the fact that limit of a product is the product of the limits. The numerator,

> Num := numer(f3(x,y));

```
2    2
Num := x  + y  - 2 x - 2 y
```

 and (1/denom) is,

> Idom := 1/denom(f3(x,y));

```
1
Idom := -----------------------
2    2
x  + y  - 2 x + 2 y + 2
```

 Notice that the denominator can be written as:

> (x-1)^2+(y+1)^2 = expand((x-1)^2+(y+1)^2);

```
2          2    2    2
(x - 1)  + (y + 1)  = x  + y  - 2 x + 2 y + 2
```

 so that it is NEVER negative for any values of x and y. The limit of Idom clearly goes to infinity as (x,y) -> (1,-1). Just take delta = 1/sqrt(M) for any large M.

Example 4

The equation of the general line towards [x4,y4,z4] is,

> L := evalm([x4,y4,z4] + t*[m1,m2,m3]);

```
L := [ 2 + t m1, 3 + t m2, t m3 ]
```

> Limit(f4(L[1],L[2],L[3]),t=0) = limit(f4(L[1],L[2],L[3]),t=0);
```
Limit  (2 + t m1) exp(t m3) + ln(1 + 2 t m1 - t m2) = 2
t -> 0
```

 This function is continuous at this point,

> f4(x4,y4,z4);

```
2
```

So the limit is just the value of the function there, i.e. 2.

Example 5

> Limit(f5(x,m*x),x=x5) = limit(f5(x,m*x),x=x5);

```
3    3  2
x  + x  m
Limit  ---------- = 0
x -> 0  2    2  2
x  + m  x
```

> Limit(f5(x,x^2),x=x5) = limit(f5(x,x^2),x=x5);
```
3    5
x  + x
Limit  ------- = 0
x -> 0  2    4
x  + x
```

> Limit(f5(y^2,y),y=y5) = limit(f5(y^2,y),y=y5);
```
6    4
y  + y
Limit  ------- = 0
y -> 0  4    2
y  + y
```

 OK good candidate for a limit. In fact just notice that

> f5(x,y);

```
3      2
x  + x y
---------
2    2
x  + y
```

> simplify(");
```
x
```

 So of course the limit exists and it is 0.

Link to the commands in this file
Carlos Rodriguez <carlos@math.albany.edu>