$f(x,y)=ln(\frac{19x^2-x^2y^2+19y^2}{x^2+y^2})$ is given as continuous function, so there exist $lim_{(x,y)\rightarrow(0,0)}f(x,y)$ and it is equal to f(0,0).

Put x=rcosA annd y=rsinA

$f(r,A)=ln(\frac{19r^2cos^2A-r^2cos^2A*r^2sin^2A+19r^2sin^2A}{r^cos^2A+r^2sin^2A})=ln(\frac{19r^2(cos^2A+sin^2A)-r^4cos^2Asin^a}{r^2(cos^2A+sin^2A)})$

we know that $cos^2A+sin^2A=1$ , so we have that

$f(r,A))=ln(\frac{19r^2-r^4cos^2Asin^a}{r^2})=ln(19-r^2cos^2Asin^2A)$

$lim_{(x,y)\rightarrow(0,0)}f(x,y)=lim_{r\rightarrow0}f(r,A)=ln19$

So f(0,0)=ln19.

8 0

3 years ago

True of false !!! The triangle shown below must be congruent

user100 [1]

False.

Right triangles do not have to be congruent.

6 0

2 years ago

Read 2 more answers

Hey guys I need your help on this!!

kogti [31]

Answer:

Domain: (-∞, ∞) or All Real Numbers

Range: (0, ∞)

Asymptote: y = 0

As x ⇒ -∞, f(x) ⇒ 0

As x ⇒ ∞, f(x) ⇒ ∞

Step-by-step explanation:

The domain is talking about the x values, so where is x defined on this graph? That would be from -∞ to ∞, since the graph goes infinitely in both directions.

The range is from 0 to ∞. This where all values of y are defined.

An asymptote is where the graph cannot cross a certain point/invisible line. A y = 0, this is the case because it is infinitely approaching zero, without actually crossing. At first, I thought that x = 2 would also be an asymptote, but it is not, since it is at more of an angle, and if you graphed it further, you could see that it passes through 2.

The last two questions are somewhat easy. It is basically combining the domain and range. However, I like to label the graph the picture attached to help even more.

As x ⇒ -∞, f(x) ⇒ 0

As x ⇒ ∞, f(x) ⇒ ∞

7 0

2 years ago