Observe that the function g is defined for all (x, y)∈R² It is easy to check, as in the above examples, that g is continuous at all point (x, y) ≠ (0, 0). Next, we shall check the continuity of g at (0, 0). For that we see if g has a limit L at (0, 0) and if L = g(0, 0) = 0. So we consider
\(\left| g\left( x,y \right) -g\left( 0,0 \right) \right| =\left| \frac { 2{ x }^{ 2 }y }{ { x }^{ 2 }+{ y }^{ 2 } } -0 \right| =\frac { 2\left| { x }^{ 2 }y \right| }{ \left| { x }^{ 2 }+{ y }^{ 2 } \right| } =\frac { 2\left| xy \right| \left| x \right| }{ { x }^{ 2 }+{ y }^{ 2 } } \le \frac { \left( { x }^{ 2 }+{ y }^{ 2 } \right) \left| x \right| }{ { x }^{ 2 }+{ y }^{ 2 } } \le \left| x \right| \)   ...(9)
Note that in the final step above we have used 2 \(\left| xy \right| \) \(\le \) x² + y² (which follows by considering 0\(\le \) (x - y)² for all  x, y∈ R . Note that (x, y)→(0, 0) implies |x| → 0. Then from (9) it follows that \(\underset { (x,y)\longrightarrow (0,0) }{ lim } \) \(\frac { 2{ x }^{ 2 }y }{ { x }^{ 2 }+{ y }^{ 2 } } \) = 0 = g (0, 0) which proves that g is continuous at (0, 0). So g is continuous at every point of R²

Note that f is defined for every (x, y)∈R². First let us check the continuity at (a, b) ≠ (0, 0).
Let us say, just for instance, (a, b) = (2, 5). Then f(2, 5) = \(\frac{10}{29}\). Then, as in the above example, we callculate \(\underset { (x,y)\longrightarrow (2,5) }{ lim } \) xy = 2(5) and \(\underset { (x,y)\longrightarrow (2,5) }{ lim } \)  x²+y² = 2²+5² = 29 ≠ 0.
Hence,  \(\underset { (x,y)\longrightarrow (2,5) }{ lim } \) \(\frac { xy }{ { x }^{ 2 }+{ y }^{ 2 } } \) = \(\frac { 10 }{ 29 } \).
Since f(2,5) = \(\frac { 10 }{ 29 } \) \(\underset { (x,y)\longrightarrow (2,5) }{ lim } \) \(\frac { xy }{ { x }^{ 2 }+{ y }^{ 2 } } \) it follows that f is continuous at (2, 5)
Exactly by similar arguments we can show that f is continuous at every point (a, b) ≠ (0, 0). Now let us check the continuity at (0, 0). Note that f (0, 0) = 0 by definition. Next we want to find if \(\underset { (x,y)\longrightarrow (0,0) }{ lim } \) \(\frac { xy }{ { x }^{ 2 }+{ y }^{ 2 } } \) exists or not.
First let us check the limit along the straight lines y = mx , passing through (0,0) .
 \(\underset { (x,y)\longrightarrow (0,0) }{ lim } \) \(\frac { xy }{ { x }^{ 2 }+{ y }^{ 2 } } \)  = \(\underset { x\longrightarrow 0 }{ lim } \) \(\frac { m{ x }^{ 2 } }{ \left( 1+{ m }^{ 2 } \right) { x }^{ 2 } } =\frac { m }{ 1+{ m }^{ 2 } } \neq \) f (0, 0), if m ≠0.
So for different values of m, we get different values \(\frac { m }{ 1+{ m }^{ 2 } } \) and hence we conclude that   \(\underset { (x,y)\longrightarrow (0,0) }{ lim } \) \(\frac { xy }{ { x }^{ 2 }+{ y }^{ 2 } } \) does not exist. Hence f cannot be continuous at (0, 0).

Volume of sphere = \(\frac43\)πr²
Given r = 10 cm 
\(\frac{dr}{dt}\) = - 0.2
V = \(\frac43\)πr³
Change in Volume
= \(\frac{4}{\not 3} \pi . \not 3 r^{2} \frac{d r}{d t}\)
= 4π(10)² (-0.2)
= 400 π (-0.2) = -80 πcm³
∴ Volume decreases by 80 π cm³
Surface area of sphere = 4πr²
Change 10 surrace area = 4 π2r\(\frac{dr}{dt}\)
= 8π(10) (-0.2)
= -\(\frac{80π\times2}{10}\) = -16 π cm²
∴ Surface area decreases by 16 π cm²

Given \(g(x)=\sqrt { { x }^{ 2 }+9 } ,{ x }_{ 0 }=-4\)
\(g(x)=\sqrt { { (-4) }^{ 2 }+9 } =\sqrt { 16+9 } =5\)
\({ g }^{ ' }(x)=\frac { 1 }{ 2 } ({ { x }^{ 2 }+9 })^{ -\frac { 1 }{ 2 } }(2x)=\frac { x }{ \sqrt { { x }^{ 2 }+9 } } \)
\(\therefore { g }^{ ' }({ x }_{ 0 })=\frac { -4 }{ \sqrt { { (-4) }^{ 2 }+9 } } =\frac { -4 }{ 5 } \)
∴ L(x) = g(x_o) + g'(x₀)(x - x_o)
= \(5-\frac { 4 }{ 5 } (x+4)=\frac { 25-4x-16 }{ 5 } \)
L(x) = \(\frac { 9-4x }{ 5 } \)

We need to find an approximate value of \(\sqrt { 9.2 } \) using linear approximation. Now by (3), we have f(x₀+Δx) ≈ f(x₀)+f'(x₀)Δx. To do this, we have to identify an appropriate function f, a point x₀ and Δx. Our choice should be such that the right side of the above approximate equality, should be computable without the help of a calculator. So, we choose  
f(x) = \(\sqrt { x,{ x }_{ 0 } } \) = 9  and Δx = 0.2. Then f'(x₀) = \(\frac { 1 }{ 2\sqrt { 9 } } \) and hence.
 \(\sqrt { 9.2 } \) ≈ f(9) + f'(9)(0.2) = 3+\(\frac { 0.2 }{ 6 } \) = 3.03333
Now if we use a calculator, just to compare, we find \(\sqrt { 9.2 } \) = 3.03315. We see that our approximation is accurate to three decimal places and the error is 3.03315 - 3.03333 = 0.00018. [Also note that one could choose f (x) = \(\sqrt { 1+x,{ x }_{ 0 } } =8\) and Δx = 0.2. So the choice of f and x₀ - are not necessarily unique].
 So in the above example, the absolute error is 3.03315-3.03333 = -0.00018. Note that the absolute error says how much the error; but it does not say how good the approximation is. For instance, let us consider two simple cases
Case 1 : Suppose that the actual value of something is 5 and its approximated value is 4, then the absolute error is 5 − 4 = 1.
Case 2 : Suppose that the actual value of something is 100 and its approximated value is 95. In this case, the absolute error is 100 − 95 = 5. So the absolute error in the first case is smaller when compared to the second case.
Among these two approximations, which is a better approximation; and why? The absolute error does not give a clear picture about whether an approximation is a good one or not. On the other hand, if we calculate relative error or percentage of error (defined below), it will be easy to see how good an approximation is. If the actual value is zero, then we do know how close our approximate answer is to the actual value. So if the actual value is not zero.

We compute
F(\(\lambda \)x, \(\lambda \)y) = \(\frac { { \left( { \lambda x } \right) }^{ 2 }+5\left( \lambda x \right) \left( \lambda y \right) -10(\lambda y)^2 }{ 3\lambda x+7\lambda y } =\frac { { \lambda }^{ 2 } }{ \lambda } \left( \frac { { x }^{ 2 }+5xy-10{ y }^{ 2 } }{ 3x+7y } \right) =\lambda F(x,y)\)
for all \(\lambda \) ∈ R. So F is a homogeneous function of degree 1.

x = 2 and dx = 0.1
Taking differentials,
df = (2x + 3) dx
Whenx = 2, dx = 0.1
df = (2(2) + 3)(0.1) = 7(0.1) = 0.7

Let F(x, y) = x² – 2y² + 2xy and x(t) = cost, y(t) = sint
Then F(x, y) = cos² t - 2sin² t + 2cos t sin t  and thus F has becomes a function of one variable t. So by using chain rule, we see that
\(\frac { dF }{ dt } \) = 2 cos t(-sin t) -4 sin t cos t 2 (-sin² t + cos² t).
= -6 cos t sin t +2 ( -sin² t + cos² t)
On the other hand if we calculate
\(\frac { \partial F }{ \partial x } \frac { \partial x }{ \partial t } +\frac { \partial F }{ \partial y } \frac { \partial y }{ \partial t } \) = (2x+2y)\(\frac { d x }{ dt } \)+(2x - 4y) \(\frac { dy }{ dt } \)
= 2(cos t + sin t)(-sin t) + 2(cos t - 2sin t)(cos t)
= -6 cos t sin t +2 ( -sin² t + cos² t)
= \(\frac { dF }{ dt } \)

We need to show that u satisfies the Laplace’s equation in R². Observe that u_x(x, y) = e^-2y(-2)sin(2x) and hence u_xx (x, y) = e^-2y(-2)(2) cos(2x).
Similarly, u_y( x y) = e^-2y (-2)cos(2x) and u_yy (x, y) = (-2)(-2)e^-2ycos(2x)
Thus, u_xx + u_yy = -4e^-2y cos(2x) + 4e^-2y cos(2x) = 0.