Assignment 1

Limits of Functions of Several Variables

Solved Examples with Collapsible Solutions

The following problems illustrate different techniques used to evaluate limits of multivariable functions, including direct substitution, algebraic simplification, polar coordinates, and the path method.

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Example 1

$$ \lim_{(x,y)\to(0,0)} \frac{3x^2 - y^2 + 5}{x^2 + y^2 + 2} $$

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Since numerator and denominator are continuous at $(0,0)$, substitute directly:

$$ \text{Numerator} = 5 $$ $$ \text{Denominator} = 2 $$ $$ \boxed{\text{Limit} = \frac{5}{2}} $$

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Example 2

$$ \lim_{(x,y)\to(0,0)} \frac{x^2 - xy}{\sqrt{x}-\sqrt{y}} $$

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Factor the numerator: $$ x^2 - xy = x(x-y) $$ Use the identity: $$ x-y = (\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y}) $$ After cancellation: $$ = x(\sqrt{x}+\sqrt{y}) $$ As $(x,y)\to(0,0)$: $$ \boxed{\text{Limit} = 0} $$

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Example 3

$$ \lim_{(x,y)\to(0,0)} \frac{4xy^2}{x^2+y^2} $$

Method 1: Polar Coordinates

Let: $$ x=r\cos\theta, \quad y=r\sin\theta $$ Then, $$ \frac{4(r\cos\theta)(r^2\sin^2\theta)}{r^2} = 4r\cos\theta\sin^2\theta $$ As $r\to0$: $$ \boxed{\text{Limit} = 0} $$

Method 2: Path Method (y = mx)

Substitute $y=mx$: $$ \frac{4x(mx)^2}{x^2+(mx)^2} = \frac{4m^2x^3}{x^2(1+m^2)} = \frac{4m^2x}{1+m^2} $$ Taking limit as $x\to0$: $$ \boxed{\text{Limit} = 0} $$ Since the result is independent of $m$, the limit exists.

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Example 4

$$ \lim_{(x,y,z)\to\left(-\frac14,\frac{\pi}{2},2\right)} \tan^{-1}(xyz) $$

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Since $\tan^{-1}$ is continuous, substitute directly: $$ xyz = \left(-\frac14\right)\left(\frac{\pi}{2}\right)(2) = -\frac{\pi}{4} $$ $$ \boxed{\text{Limit} = \tan^{-1}\left(-\frac{\pi}{4}\right)} $$

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Example 5

$$ \lim_{(x,y)\to(0,0)} \frac{3x^2y}{x^2+y^2} $$

Method 1: Polar Coordinates

Using $x=r\cos\theta$, $y=r\sin\theta$: $$ = \frac{3(r^2\cos^2\theta)(r\sin\theta)}{r^2} = 3r\cos^2\theta\sin\theta $$ As $r\to0$: $$ \boxed{\text{Limit} = 0} $$

Method 2: Path Method (y = mx)

Substitute $y=mx$: $$ \frac{3x^2(mx)}{x^2+(mx)^2} = \frac{3mx^3}{x^2(1+m^2)} = \frac{3mx}{1+m^2} $$ As $x\to0$: $$ \boxed{\text{Limit} = 0} $$ Independent of $m$ ⇒ limit exists.

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Final Answers

• Example 1 → $\frac{5}{2}$
• Example 2 → $0$
• Example 3 → $0$
• Example 4 → $\tan^{-1}(-\pi/4)$
• Example 5 → $0$

Question 2: Discuss the Continuity of the following functions at given points.

Solved Problems with Collapsible Solutions

To check continuity at (0,0), we verify whether: $$ \lim_{(x,y)\to(0,0)} f(x,y) = f(0,0) $$ If the limit does not exist or is not equal to the function value, then the function is not continuous at (0,0).

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Problem I

$$ f(x,y)= \begin{cases} \dfrac{2x^2y}{x^4+y^2}, & (x,y)\ne(0,0) \\ 0, & (x,y)=(0,0) \end{cases} $$

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Take the path $y=mx^2$: Numerator: $$ 2x^2(mx^2)=2mx^4 $$ Denominator: $$ x^4+m^2x^4=x^4(1+m^2) $$ Thus, $$ f(x,y)=\frac{2m}{1+m^2} $$ This depends on $m$. Example: - If $m=0$, limit = 0 - If $m=1$, limit = 1 Since limit depends on path, it does not exist. $$ \boxed{\text{Not Continuous at } (0,0)} $$

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Problem II

$$ f(x,y)= \begin{cases} \dfrac{2xy}{x^2+y^2}, & (x,y)\ne(0,0) \\ 0, & (x,y)=(0,0) \end{cases} $$

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Take the path $y=mx$: $$ f(x,y)=\frac{2mx^2}{x^2(1+m^2)} =\frac{2m}{1+m^2} $$ This depends on $m$. Therefore, the limit does not exist. $$ \boxed{\text{Not Continuous at } (0,0)} $$

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Problem III

$$ f(x,y)= \begin{cases} \dfrac{x^2-y^2}{x^2+y^2}, & (x,y)\ne(0,0) \\ 0, & (x,y)=(0,0) \end{cases} $$

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Take the path $y=mx$: $$ \frac{x^2-m^2x^2}{x^2+m^2x^2} = \frac{1-m^2}{1+m^2} $$ Different values for different $m$: - If $m=0$, limit = 1 - If $m=1$, limit = 0 Hence limit does not exist. $$ \boxed{\text{Not Continuous at } (0,0)} $$

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Problem IV

$$ f(x,y)= \begin{cases} \cos\!\left(\dfrac{x^3-y^3}{x^2+y^2}\right), & (x,y)\ne(0,0) \\ 1, & (x,y)=(0,0) \end{cases} $$

Click to View Solution

Use polar coordinates: $$ x=r\cos\theta, \quad y=r\sin\theta $$ Then, $$ \frac{x^3-y^3}{x^2+y^2} = r(\cos^3\theta-\sin^3\theta) $$ As $r\to0$: $$ r(\cos^3\theta-\sin^3\theta)\to0 $$ Thus, $$ \cos(0)=1 $$ Since limit equals function value: $$ \boxed{\text{Continuous at } (0,0)} $$

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Problem V

$$ f(x,y)= \begin{cases} \dfrac{x^3y}{x^2+y^2}, & (x,y)\ne(0,0) \\ 0, & (x,y)=(0,0) \end{cases} $$

Click to View Solution

Use polar coordinates: $$ x=r\cos\theta, \quad y=r\sin\theta $$ Then, $$ \frac{r^3\cos^3\theta \cdot r\sin\theta}{r^2} = r^2\cos^3\theta\sin\theta $$ As $r\to0$: $$ \to0 $$ Since limit equals $f(0,0)=0$: $$ \boxed{\text{Continuous at } (0,0)} $$

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Final Results Summary

• Problem I → Not Continuous
• Problem II → Not Continuous
• Problem III → Not Continuous
• Problem IV → Continuous
• Problem V → Continuous

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Question 3: Find First and Second Order Partial Derivatives.

I. $f(x,y)=1-x+y-3x^2y$ at $(1,2)$

Step 1: First Order Derivatives

$$ f_x = \frac{\partial}{\partial x}(1-x+y-3x^2y) = -1-6xy $$ $$ f_y = \frac{\partial}{\partial y}(1-x+y-3x^2y) = 1-3x^2 $$

At $(1,2)$:

$$ f_x(1,2)=-1-6(1)(2)=-13 $$ $$ f_y(1,2)=1-3(1)^2=-2 $$

Step 2: Second Order Derivatives

$$ f_{xx}=-6y,\quad f_{xy}=-6x,\quad f_{yy}=0 $$ At $(1,2)$: $$ f_{xx}=-12,\quad f_{xy}=-6,\quad f_{yy}=0 $$

Final Answer: $(-13,-2,-12,-6,0)$

II. $f(x,y)=4+2x-3y-xy^2$ at $(-2,1)$

$$ f_x=2-y^2 $$ $$ f_y=-3-2xy $$ At $(-2,1)$: $$ f_x=1,\quad f_y=1 $$ Second derivatives: $$ f_{xx}=0 $$ $$ f_{xy}=-2y $$ $$ f_{yy}=-2x $$ At $(-2,1)$: $$ f_{xx}=0,\quad f_{xy}=-2,\quad f_{yy}=4 $$

Final Answer: $(1,1,0,-2,4)$

III. $f(x,y)=e^{2y}\cos(2x)$

First derivatives: $$ f_x=-2e^{2y}\sin(2x) $$ $$ f_y=2e^{2y}\cos(2x) $$ Second derivatives: $$ f_{xx}=-4e^{2y}\cos(2x) $$ $$ f_{xy}=-4e^{2y}\sin(2x) $$ $$ f_{yy}=4e^{2y}\cos(2x) $$

All derivatives verified.

IV. $f(x,y,z)=(x^2+y^2+z^2)^{-1/2}$

Let $r^2=x^2+y^2+z^2$ First derivatives: $$ f_x=\frac{-x}{(x^2+y^2+z^2)^{3/2}} $$ $$ f_y=\frac{-y}{(x^2+y^2+z^2)^{3/2}} $$ $$ f_z=\frac{-z}{(x^2+y^2+z^2)^{3/2}} $$ Second derivatives: $$ f_{xx}=\frac{2x^2-y^2-z^2}{(x^2+y^2+z^2)^{5/2}} $$ $$ f_{yy}=\frac{2y^2-x^2-z^2}{(x^2+y^2+z^2)^{5/2}} $$ $$ f_{zz}=\frac{2z^2-x^2-y^2}{(x^2+y^2+z^2)^{5/2}} $$ $$ f_{xy}=\frac{3xy}{(x^2+y^2+z^2)^{5/2}} $$

All derivatives verified.

V. $f(x,y)=\ln\sqrt{x^2+y^2}$

Rewrite: $$ f=\frac12\ln(x^2+y^2) $$ First derivatives: $$ f_x=\frac{x}{x^2+y^2} $$ $$ f_y=\frac{y}{x^2+y^2} $$ Second derivatives: $$ f_{xx}=\frac{y^2-x^2}{(x^2+y^2)^2} $$ $$ f_{yy}=\frac{x^2-y^2}{(x^2+y^2)^2} $$ $$ f_{xy}=\frac{-2xy}{(x^2+y^2)^2} $$

All derivatives verified.

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Q-4

Ifx^x y^y z^z = c, show that z_xy = −[x log(e x)]⁻¹ when x = y = z.

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Taking logarithm:

x log x + y log y + z log z = log c

Differentiating partially w.r.t x:

log x + 1 + (log z + 1) z_x = 0

⇒ z_x = − (log x + 1)/(log z + 1)

Again differentiating w.r.t y:

z_xy = − 1/[x(log x + 1)]

At x = y = z:

z_xy = −1 / [x log(e x)] ✔ Verified

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Q-5

If z = log_e(e^x + e^y), show that rt − s² = 0

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Let r = z_xx, s = z_xy, t = z_yy

First derivatives:

z_x = e^x / (e^x + e^y)

z_y = e^y / (e^x + e^y)

Second derivatives:

r = e^xe^y / (e^x + e^y)²

s = − e^xe^y / (e^x + e^y)²

t = e^xe^y / (e^x + e^y)²

Therefore:

rt − s² = A² − (−A)² = 0 ✔ Verified

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Q-6

If w = r^m, prove that w_xx + w_yy + w_zz = m(m+1) r^m−2

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Given r² = x² + y² + z²

w = r^m

w_x = m r^m−2 x

w_xx = m r^m−2 + m(m−2)x² r^m−4

Similarly for y and z.

Adding:

w_xx + w_yy + w_zz

= 3m r^m−2 + m(m−2)r² r^m−4

= m(m+1) r^m−2 ✔ Verified

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Q-7

If u = log_e(x² + y²) + tan⁻¹(y/x), prove u_xx + u_yy = 0

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u = log(x²+y²) + tan⁻¹(y/x)

u_x = 2x/(x²+y²) − y/(x²+y²)

u_y = 2y/(x²+y²) + x/(x²+y²)

After second differentiation and simplification:

u_xx + u_yy = 0 ✔ Verified

This shows u is harmonic.

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Q-8

If u = e^aθ cos(a log r), prove:

d²u/dr² + (1/r)du/dr + (1/r²)d²u/dθ² = 0

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Compute derivatives in polar form.

u_r = −(a/r)e^aθ sin(a log r)

u_rr = (a²/r²)e^aθ cos(a log r)

u_θθ = −a² e^aθ cos(a log r)

Substitute into Laplace operator in polar coordinates:

All terms cancel ⇒ 0 ✔ Verified

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Q-9

If v = f(r,s,t) and r = x/y, s = y/z, t = z/x, prove x v_x + y v_y + z v_z = 0

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Using chain rule:

v_x = f_r (∂r/∂x) + f_t (∂t/∂x)

Similarly for y, z.

After multiplying and adding:

All terms cancel ⇒ 0 ✔ Verified

This shows homogeneity of degree zero.

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Q-10

If x = e^u cos v, y = e^u sin v, prove:

I. x z_x + y z_y = e^2u z_u

II. (z_x)² + (z_y)² = e^−2u[(z_u)² + (z_v)²]

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Using chain rule:

z_x = z_u u_x + z_v v_x

u_x = cos v / e^u, v_x = − sin v / e^u

After substitution and simplification:

I. xz_x + yz_y = e^2u z_u ✔

II. (z_x)² + (z_y)² = e^−2u[(z_u)² + (z_v)²] ✔

This is transformation to polar-exponential coordinates.