Exercise 8.3 - Chapter 8 Differentials and Partial Derivatives 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $8.3$
Question 1.
Evaluate $\lim _{(x, y) \rightarrow(1,2)} g(x, y)$, if the limit exists, where $g(x, y)=\frac{3 x^2-x y}{x^2+y^2+3}$.
Solution:
$
\begin{aligned}
& \lim _{(x, y) \rightarrow(1,2)} g(x, y) \\
& =(x, y) \rightarrow(1,2) \frac{3 x^2-x y}{x^2+y^2+3} \\
& =\frac{\lim _{(x, y) \rightarrow(1,2)} 3 x^2-x y}{\lim _{(x, y) \rightarrow(1,2)} x^2+y^2+3}=\frac{3(1)^2-(1)(2)}{(1)^2+(2)^2+3} \\
& =\frac{3-2}{1+4+3}=\frac{1}{8}
\end{aligned}
$
If $\lim _{(x, y) \rightarrow(a, b)} f(x, y)=f(a, b)$ then the limit exists.
Question 2.
Evaluate $\lim _{(x, y) \rightarrow(0,0)} \cos \left(\frac{x^3+y^2}{x+y+2}\right)$. If the limit exists.
Solution:
$
\lim _{(x, y) \rightarrow(0,0)} \cos \left(\frac{x^3+y^2}{x+y+2}\right)
$
Let us take $y=0$ for $f(y)$, then $\lim _{y \rightarrow 0} \cos \left(\frac{y^2}{y+2}\right)=1 \neq \infty$
$\therefore$ limit exists
Now, $\lim _{(x, y) \rightarrow(0,0)} \cos \left(\frac{x^3+y^2}{x+y+2}\right)=\cos \left(\frac{0}{2}\right)=\cos (0)=1$

Question 3.
Let $f(x, y)=\frac{y^2-x y}{\sqrt{x}-\sqrt{y}}$ for $(x, y) \neq(0,0)$. Show that $\lim _{(x, y) \rightarrow(0,0)} f(x, y)=0$.
Solution:
$
\begin{aligned}
\lim _{(x, y) \rightarrow(0,0)} f(x, y) & =\lim _{(x, y) \rightarrow(0,0)} \frac{y^2-x y}{\sqrt{x}-\sqrt{y}} \\
& =\lim _{(x, y) \rightarrow(0,0)} \frac{y^2-x y}{\sqrt{x}-\sqrt{y}} \times \frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}+\sqrt{y}} \\
& =\lim _{(x, y) \rightarrow(0,0)} \frac{-y(x-y)(\sqrt{x}+\sqrt{y})}{(x-y)} \\
& =\lim _{(x, y) \rightarrow(0,0)}-y(\sqrt{x}+\sqrt{y})=0
\end{aligned}
$
Hence proved

Question 4.
Evaluate $\lim _{(x, y) \rightarrow(0,0)} \cos \left(\frac{\mathrm{e}^x \sin y}{y}\right)$, if the limit exists.
Solution:
$
(x, y) \rightarrow(0,0) \cos \left(\frac{e^x \sin y}{y}\right)
$
Using two different parts to approach the point along $x=0$ and $y=0$
$
\begin{aligned}
& \lim _{(x, y) \rightarrow(0,0)} \cos \left(\frac{e^x \sin y}{y}\right) \\
& =\cos \left(\lim _{(x, y) \rightarrow(0,0)} \frac{e^x \sin y}{y}\right)=\operatorname{Cos}(1) \quad\left(\because \lim _{y \rightarrow 0} \frac{\sin y}{y}=1\right)
\end{aligned}
$

Question 5.
Let $g(x, y)=\frac{x^2 y}{x^4+y^2}$ for $(x, y) \neq(0,0)$ and $f(0,0)=0$.
(i) Show that $\lim _{(x, y) \rightarrow(0,0)} g(x, y)=0$ along every line $y=m x, m \in R$.
(ii) Show that $\lim _{(x, y) \rightarrow(0,0)} g(x, y)=\frac{k}{1+k^2}$ along every parabola $y=k x^2, k \in \mathbf{R} \backslash\{0\}$.
Solution:
Given $g(x, y)=\frac{x^2 y}{x^4+y^2}$ for $(x, y) \neq(0,0)$ and $f(0,0)=0$
(i) Along the line $y=m x, m \in \mathrm{R}$
$
\begin{aligned}
& \lim _{(x, y) \rightarrow(0,0)} g(x, y)=\lim _{(x, m x) \rightarrow(0,0)}\left(\frac{x^2(m x)}{x^4+(m x)^2}\right) \\
& =\lim _{(x, m x) \rightarrow(0,0)}\left(\frac{m x^3}{x^4+m^2 x^2}\right) \\
& =\lim _{(x, m x) \rightarrow(0,0)}\left(\frac{m x}{x^2+m^2}\right)=\frac{0}{m^2}=0 \\
&
\end{aligned}
$

Hence Proved.

(ii) $\begin{aligned}
g(x, y) & =\frac{x^2 y}{x^4+y^2} \text { along every parabola } y=k x^2, k \in \mathbb{R} \backslash\{0\} \\
\lim _{(x, y) \rightarrow(0,0)} g(x, y) & =\lim _{\left(x, k x^2\right) \rightarrow(0,0)} \frac{x^2\left(k x^2\right)}{x^4+\left(k x^2\right)^2} \\
& =\left(x, k x^2\right) \rightarrow(0,0) \frac{k x^4}{x^4\left(1+k^2\right)} \\
& =\left(x, k x^2\right) \rightarrow(0,0)\left(\frac{k}{1+k^2}\right)=\frac{k}{1+k^2}
\end{aligned}$

Hence Proved.

Question 6.
Show that $f(x, y)=\frac{x^2-y^2}{y^2+1}$ is continuous at every $(x, y) \in \mathbf{R}^2$.
Solution:
$
f(x, y)=\frac{x^2-y^2}{y^2+1}
$
Let $(a, b) \in \mathbb{R}^2$ be an arbitrary point.
Now $f(a, b)=\frac{a^2-b^2}{b^2+1}$ is defined
$
\begin{aligned}
& \lim _{(x, y) \rightarrow(a, b)} \frac{x^2-y^2}{y^2+1}=\frac{\lim _{(x, y) \rightarrow(a, b)}\left[x^2-y^2\right]}{\lim _{(x, y) \rightarrow(a, b)}\left[y^2+1\right]}=\frac{a^2-b^2}{b^2+1}=\mathrm{L} \quad \text { limit exists } \\
& \lim _{(x, y) \rightarrow(a, b)} f(x, y)=\mathrm{L}=f(a, b) \\
&
\end{aligned}
$
Here, $f$ satisfies all the three conditions of continuity at (a, b). Hence, $f^{\prime}$ is continuous at every point of $\mathrm{R}^2$ as $(\mathrm{a}, \mathrm{b}) \in \mathrm{R}^2$.

Question 7.
Let $g(x, y)=\frac{e^y \sin x}{x}$, for $x \neq 0$ and $g(0,0)=1$. Show that $g$ is continuous at $(0,0)$.
Solution:
$g(x, y)=\frac{e^y \sin x}{x}, x \neq 0$ and $g(0,0)=1$
By definition $g(0,0)=1$ (given)
(i.e.,)
$
\begin{aligned}
& \lim _{(x, y) \rightarrow(0,0)} g(x, y)=\lim _{(x, y) \rightarrow(0,0)} \frac{e^y \sin x}{x} \\
& =\left[\lim _{(x, y) \rightarrow(0,0)} e^y\right]\left[\begin{array}{l}
x \\
\lim _{(x, y) \rightarrow(0,0)} \frac{\sin x}{x}
\end{array}\right] \\
& =(1)(1)=1 \\
&
\end{aligned}
$
Now, limit along the line $y=m x$, passing through $(0,0)$
$
\lim _{(x, m x)} \frac{e^{m x} \sin x}{x}=1 \text { (defined) }
$
Hence, ' $g$ ' is continuous at $(0,0)$