# By considering different path of approach, show that the function f(x,y)=(x^2+y)/y has no limis (x,y) ->(0,0) Question
Multivariable functions By considering different path of approach, show that the function $$\displaystyle{f{{\left({x},{y}\right)}}}=\frac{{{x}^{{2}}+{y}}}{{y}}$$ has no limis $$\displaystyle{\left({x},{y}\right)}\to{\left({0},{0}\right)}$$ 2021-01-03
It is known that when finding the limits of the multivariable function we have to find the limits using different paths .
Take the path along the x-axis ,then $$\displaystyle{\left({x},{y}\right)}\to{\left({x},{0}\right)}$$
$$\displaystyle\lim_{{{\left({x},{y}\right)}\to{\left({x},{0}\right)}}}{f{{\left({x}\right)}}}\lim{\left({\left({x},{y}\right)}\to{\left({x},{0}\right)}\right)}\frac{{{x}^{{2}}+{y}}}{{y}}={0}$$
Take the path along the y-axis ,then
$$\displaystyle\lim_{{{\left({x},{y}\right)}\to{\left({x},{0}\right)}}}{f{{\left({x}\right)}}}\lim{\left({\left({x},{y}\right)}\to{\left({x},{0}\right)}\right)}\frac{{{x}^{{2}}+{y}}}{{y}}={1}$$
Using the different paths , the values of the limits is different , then the limit does not exist.
Thus, the given function $$\displaystyle{f{{\left({x},{y}\right)}}}=\frac{{{x}^{{2}}+{y}}}{{y}}$$ has no limits

### Relevant Questions Consider this multivariable function. f(x,y)=xy+2x+y−36
a) What is the value of f(2,−3)?
b) Find all x-values such that f (x,x) = 0 A surface is represented by the following multivariable function,
$$\displaystyle{f{{\left({x},{y}\right)}}}=\frac{{1}}{{3}}{x}^{{3}}+{y}^{{2}}-{2}{x}{y}-{6}{x}-{3}{y}+{4}$$
a) Calculate $$\displaystyle{f}_{{\times}},{f}_{{{y}{x}}},{f}_{{{x}{y}}}{\quad\text{and}\quad}{f}_{{{y}{y}}}$$
b) Calculate coordinates of stationary points.
c) Classify all stationary points. Consider this multivariable function. $$\displaystyle{f{{\left({x},{y}\right)}}}={y}{e}^{{{3}{x}}}+{y}^{{2}}$$
a) Find $$\displaystyle{{f}_{{y}}{\left({x},{y}\right)}}$$
b) What is value of $$\displaystyle{{f}_{{\times}}{\left({0},{3}\right)}}$$? Comsider this, multivariable fucntion
$$\displaystyle{f{{\left({x},{y}\right)}}}={3}{x}{y}-{x}^{{2}}+{5}{y}^{{2}}-{25}$$
a)What is the value of f(-1,3)?
b)Find all x-values such that f(x,x)=0 A surface is represented by the following multivariable function,
$$\displaystyle{f{{\left({x},{y}\right)}}}={x}^{{3}}+{y}^{{3}}-{3}{x}-{3}{y}+{1}$$
Calculate coordinates of stationary points. Average value over a multivariable function using triple integrals. Find the average value of $$\displaystyle{F}{\left({x},{y},{z}\right)}={x}^{{2}}+{y}^{{2}}+{z}^{{2}}$$ over the cube in the first octant bounded bt the coordinate planes and the planes x=5, y=5, and z=5 Write formulas for the indicated partial derivatives for the multivariable function.
$$\displaystyle{f{{\left({x},{y}\right)}}}={7}{x}^{{2}}+{9}{x}{y}+{4}{y}^{{3}}$$
a)$$\displaystyle\frac{{\partial{f}}}{{\partial{x}}}$$
b)(delf)/(dely)ZSK
c)$$\displaystyle\frac{{\partial{f}}}{{\partial{x}}}{\mid}_{{{y}={9}}}$$ Let $$\displaystyle{z}={e}^{{{x}^{{2}}+{3}{y}}}+{x}^{{3}}{y}^{{2}}$$, where $$\displaystyle{x}={t}{\cos{{r}}}{\quad\text{and}\quad}{y}={r}{t}^{{4}}$$. Use the chain rule fot multivariable functions (Cals III Chain Rule) to find $$\displaystyle\frac{{\partial{z}}}{{\partial{r}}}$$ and $$\displaystyle\frac{{\partial{z}}}{\partial}{t}{)}$$. Give your answers in terms of r and t only. Be sure to show all of your work. Use polar coordinates to find the limit. [Hint: Let $$\displaystyle{x}={r}{\cos{{\quad\text{and}\quad}}}{y}={r}{\sin{}}$$ , and note that (x, y) (0, 0) implies r 0.] $$\displaystyle\lim_{{{\left({x},{y}\right)}\to{\left({0},{0}\right)}}}\frac{{{x}^{{2}}-{y}^{{2}}}}{\sqrt{{{x}^{{2}}+{y}^{{2}}}}}$$ Find the critical points of the following multivariable function, and using the Second Derivative Test to find the maxima and minima of the equation below: $$\displaystyle{a}{\left({x},{y}\right)}=\frac{{{8}{x}-{y}}}{{{e}^{{{x}^{{2}}+{y}^{{2}}}}}}$$