Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 8 Differentials and Partial Derivatives Ex 8.3 Textbook Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 8 Differentials and Partial Derivatives Ex 8.3

Question 1.
Evaluate $$\lim _{(x, y) \rightarrow(1,2)}$$ g(x, y), if the limit exists where g(x, y) = $$\frac { 3x^2-xy }{ x^2+y^2+3 }$$
Solution:
$$\lim _{(x, y) \rightarrow(1,2)}$$ g(x, y) = $$\lim _{(x, y) \rightarrow(1,2)}$$ $$\frac { 3x^2-xy }{ x^2+y^2+3 }$$
= $$\frac { 3(1)-1×2 }{ 1+4+3 }$$
= $$\frac { 1 }{ 8 }$$

Question 2.
Evaluate $$\lim _{(x, y) \rightarrow(0,0)}$$ cos($$\frac { x^3+y^2 }{ x+y+2 }$$), if the limit exists.
Solution:
$$\lim _{(x, y) \rightarrow(0,0)}$$ cos($$\frac { x^3+y^2 }{ x+y+2 }$$)
= cos ($$\frac { 0+0 }{ 0+0+2 }$$)
= cos 0
= $$\lim _{(x, y) \rightarrow(0,0)}$$ cos($$\frac { x^3+y^2 }{ x+y+2 }$$) = 1

Question 3.
Let f(x, y) $$\frac { y^2-xy }{ √x-√y }$$ for (x, y) ≠ (0, 0) show that $$\lim _{(x, y) \rightarrow(0,0)}$$ f(x, y) = 0
Solution:
f(x, y) = $$\frac { y^2-xy }{ √x-√y }$$

Question 4.
Evaluate $$\lim _{(x, y) \rightarrow(0,0)}$$ cos($$\frac { e^xsin y }{ y }$$), if the limit exists.
Solution:

Question 5.
Let g(x, y) = $$\frac { x^2y }{ x^4+y^2 }$$ for (x, y) ≠ (0, 0) and f(0, 0) = 0
(i) Show that $$\lim _{(x, y) \rightarrow(0,0)}$$ g(x, y) = 0 along every line y = mx, m ∈ R
(ii) Show that $$\lim _{(x, y) \rightarrow(0,0)}$$ g(x, y) = $$\frac { k }{ 1+k^2 }$$ along every parabola y = kx², k ∈ R\{0}
Solution:
(i) Let g(x, y) = $$\frac { x^2y }{ x^4+y^2 }$$ for (x, y) ≠ (0, 0) and f(0, 0) = 0
Let y = mx

= 0
Hence proved

(ii) for parabola y = kx²

Hence proved

Question 6.
Show that f (x, y) = $$\frac { x^2-y^2 }{ y^+1 }$$ is continuous at every (x, y) ∈ R²
Solution:
Let (a, b) ∈ R² be an arbitrary point.
We shall investigate the continuity of f at (a,b).
That is, we shall check if all the three conditions for continuity hold for f at (a, b)
To check first condition, note that
f(a, b) = $$\frac { a^2-b^2 }{ b^2+1 }$$ is defined
Next we want t0 find lf $$\lim _{(x, y) \rightarrow(a,b)}$$ f(x, y) exist or not
so we calculate $$\lim _{(x, y) \rightarrow(a,b)}$$ x² – y² = a² – b² and
$$\lim _{(x, y) \rightarrow(a,b)}$$ y² + 1 = b² + 1
By the properties of limit we see that
$$\lim _{(x, y) \rightarrow(a,b)}$$ f(x, y) = $$\frac { x^2-y^2 }{ y^2+1 }$$ = $$\frac { a^2-b^2 }{ b^2+1 }$$ = f(a, b) = L exists
Now, we note that $$\lim _{(x, y) \rightarrow(a,b)}$$ f(x, y) = L = f(a, b).
Hence f satisfies all the there conditions for continuity of f at (a, b).
Since (a, b) is an arbitrary point in R², we conclude that f is continuous at every point of R².

Question 7.
Let g (x, y) = $$\frac { e^ysinx }{ x }$$ for x ≠ 0 and g(0, 0) = 1 shoe that g is continuous at (0, 0)
Solution:
g(x, y) = $$\frac { e^ysinx }{ x }$$
To check first condition,note that
$$\lim _{(x, y) \rightarrow(0,0)}$$ g(x, y) = $$\lim _{(x, y) \rightarrow(0,0)}$$ $$\frac { e^ysinx }{ x }$$ = 1 is defined
Next $$\lim _{(x, y) \rightarrow(0,0)}$$ g(x, y) = $$\frac { e^ysinx }{ x }$$ = 1 = L Exist
Now we note $$\lim _{(x, y) \rightarrow(0,0)}$$ g(x, y) = L = g(0, 0)
Hence g satisfies all the three conditions for; continuity of g at (0, 0).
We conclude that g is continuous at (0, 0).