Samacheer Kalvi Class 12 Maths Solutions Chapter 8 Differentials and Partial Derivatives Exercise 8.3

Question 1. Evaluate \( \lim _{(x, y) \rightarrow(1,2)} g(x, y) \), if the limit exists where \( g(x, y) = \frac { 3x^2-xy }{ x^2+y^2+3} \)
Answer: To find the limit, we can directly substitute the values \( x=1 \) and \( y=2 \) into the function, since the denominator will not be zero.
\( \lim _{(x, y) \rightarrow(1,2)} g(x, y) = \lim _{(x, y) \rightarrow(1,2)} \frac { 3x^2-xy }{ x^2+y^2+3} \)
\( = \frac { 3(1)^2-(1)(2) }{ (1)^2+(2)^2+3} \)
\( = \frac { 3-2 }{ 1+4+3} \)
\( = \frac { 1 }{ 8 } \)
Direct substitution is a common way to evaluate limits for continuous functions.
In simple words: When a function is smooth and well-behaved, you can just put the numbers for x and y directly into the equation to find the limit. Here, we put \( x=1 \) and \( y=2 \) into the formula to get the answer.

🎯 Exam Tip: For rational functions (fractions with polynomials), direct substitution is valid if the denominator does not become zero at the limit point. Always check the denominator first.

Question 2. Evaluate \( \lim _{(x, y) \rightarrow(0,0)} \cos(\frac { x^3+y^2 }{ x+y+2}) \), if the limit exists.
Answer: To evaluate this limit, we can substitute \( x=0 \) and \( y=0 \) into the expression inside the cosine function, as the denominator will not be zero.
\( \lim _{(x, y) \rightarrow(0,0)} \cos(\frac { x^3+y^2 }{ x+y+2 }) \)
\( = \cos (\frac { 0^3+0^2 }{ 0+0+2 }) \)
\( = \cos (\frac { 0 }{ 2 }) \)
\( = \cos (0) \)
\( = 1 \)
The cosine function is continuous everywhere, which allows for this direct substitution.
In simple words: Just plug in \( x=0 \) and \( y=0 \) into the equation. The part inside `cos` becomes 0, and we know that `cos(0)` is 1.

🎯 Exam Tip: When evaluating limits of composite functions like \( \cos(f(x,y)) \), you can often find the limit of the inner function \( f(x,y) \) first, then apply the outer function if it's continuous.

Question 3. Let \( f(x, y) = \frac { y^2-xy }{ \sqrt{x}-\sqrt{y} } \) for \( (x, y) \neq (0, 0) \) show that \( \lim _{(x, y) \rightarrow(0,0)} f(x, y) = 0 \)
Answer: We need to simplify the expression for \( f(x, y) \) first because direct substitution would lead to an undefined form \( \frac{0}{0} \).
\( f(x, y) = \frac { y^2-xy }{ \sqrt{x}-\sqrt{y} } \)
Factor out \( y \) from the numerator:
\( = \frac { y(y-x) }{ \sqrt{x}-\sqrt{y} } \)
We know that \( y-x = - (x-y) \) and \( x-y = (\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y}) \).
So, \( y-x = -(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y}) \).
Substitute this back into the expression:
\( = \frac { y [ -(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y}) ] }{ \sqrt{x}-\sqrt{y} } \)
Cancel out \( (\sqrt{x}-\sqrt{y}) \) from numerator and denominator (since \( x \neq y \) near the limit point):
\( = -y(\sqrt{x}+\sqrt{y}) \)
Now, we can find the limit by substituting \( x=0 \) and \( y=0 \):
\( \lim _{(x, y) \rightarrow(0,0)} f(x, y) = \lim _{(x, y) \rightarrow(0,0)} -y(\sqrt{x}+\sqrt{y}) \)
\( = -(0)(\sqrt{0}+\sqrt{0}) \)
\( = 0 \cdot 0 \)
\( = 0 \)
This shows that the limit of the function as \( (x,y) \) approaches \( (0,0) \) is 0.
In simple words: The problem looks tricky because plugging in zero gives `0/0`. We can change the top part of the fraction to `y` times `(y-x)`. Then, we use a special math trick to rewrite `(y-x)` using square roots. After that, we can cancel out similar terms from the top and bottom. This leaves a simpler expression where we can safely put in `x=0` and `y=0` to get 0.

🎯 Exam Tip: When faced with an indeterminate form like \( \frac{0}{0} \), try algebraic simplification first. This often involves factoring, rationalizing, or using special identities to eliminate the problematic terms.

Question 4. Evaluate \( \lim _{(x, y) \rightarrow(0,0)} \cos(\frac { e^x \sin y }{ y }) \), if the limit exists.
Answer: We need to evaluate the limit of the expression inside the cosine function first. We can separate the limit because the individual limits exist.
Consider the inner limit: \( \lim _{(x, y) \rightarrow(0,0)} \frac { e^x \sin y }{ y } \)
We can rewrite this as: \( \lim _{x \rightarrow 0} e^x \cdot \lim _{y \rightarrow 0} \frac { \sin y }{ y } \)
We know the standard limits:
\( \lim _{x \rightarrow 0} e^x = e^0 = 1 \)
\( \lim _{y \rightarrow 0} \frac { \sin y }{ y } = 1 \)
So, the inner limit is \( 1 \cdot 1 = 1 \).
Now, substitute this result back into the cosine function:
\( \lim _{(x, y) \rightarrow(0,0)} \cos(\frac { e^x \sin y }{ y }) = \cos(1) \)
The cosine function is continuous, allowing us to find the limit of the inner expression first.
In simple words: We first find the limit of the fraction inside the `cos` part. We split it into two known limits: `e` to the power of `x` (which becomes 1 as `x` goes to 0) and `sin y` over `y` (which also becomes 1 as `y` goes to 0). So the inner part becomes 1. Then, `cos(1)` is the final answer.

🎯 Exam Tip: Remember key trigonometric limits like \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \) and properties of exponential functions, as they are frequently used in multivariable limits.

Question 5. Let \( g(x, y) = \frac { x^2y }{ x^4+y^2 } \) 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 \( y = mx, m \in R \)
(ii) Show that \( \lim _{(x, y) \rightarrow(0,0)} g(x, y) = \frac { k }{ 1+k^2 } \) along every parabola \( y = kx^2, k \in R\{0\} \)
Answer:
(i) Along every line \( y = mx \):
Substitute \( y = mx \) into the function \( g(x, y) \):
\( g(x, y) = \frac { x^2(mx) }{ x^4+(mx)^2 } \)
\( = \frac { mx^3 }{ x^4+m^2x^2 } \)
Factor out \( x^2 \) from the denominator:
\( = \frac { mx^3 }{ x^2(x^2+m^2) } \)
Cancel \( x^2 \) (for \( x \neq 0 \)):
\( = \frac { mx }{ x^2+m^2 } \)
Now take the limit as \( (x, y) \rightarrow(0,0) \), which means \( x \rightarrow 0 \):
\( \lim _{x \rightarrow 0} \frac { mx }{ x^2+m^2 } = \frac { m(0) }{ 0^2+m^2 } = \frac { 0 }{ m^2 } = 0 \) (assuming \( m \neq 0 \)).
If \( m=0 \), then \( y=0 \). In this case, \( g(x,0) = \frac{x^2(0)}{x^4+0^2} = 0 \) for \( x \neq 0 \). So the limit is still 0.
Thus, the limit is 0 along every line \( y = mx \). This step helps understand path-dependent limits.
(ii) Along every parabola \( y = kx^2 \), where \( k \in R\{0\} \):
Substitute \( y = kx^2 \) into the function \( g(x, y) \):
\( g(x, y) = \frac { x^2(kx^2) }{ x^4+(kx^2)^2 } \)
\( = \frac { kx^4 }{ x^4+k^2x^4 } \)
Factor out \( x^4 \) from the denominator:
\( = \frac { kx^4 }{ x^4(1+k^2) } \)
Cancel \( x^4 \) (for \( x \neq 0 \)):
\( = \frac { k }{ 1+k^2 } \)
Now take the limit as \( (x, y) \rightarrow(0,0) \), which means \( x \rightarrow 0 \):
\( \lim _{x \rightarrow 0} \frac { k }{ 1+k^2 } = \frac { k }{ 1+k^2 } \)
The limit along these parabolic paths depends on \( k \), which means the limit does not exist at \( (0,0) \) because it takes different values along different paths.
In simple words: This question asks us to check how the function behaves when we approach `(0,0)` along different paths.
(i) When we approach along straight lines (`y = mx`), the answer for the limit is always 0.
(ii) But when we approach along curved paths like parabolas (`y = kx^2`), the answer for the limit changes depending on the value of `k`. Because the limit is not the same for all paths, it means the overall limit at `(0,0)` does not exist, even though the question asks to show a specific value for these paths.

🎯 Exam Tip: To prove a limit does not exist, show that the function approaches different values along at least two different paths to the point. If the limit exists, it must be the same along all possible paths.

Question 6. Show that \( f(x, y) = \frac { x^2-y^2 }{ y^2+1 } \) is continuous at every \( (x, y) \in R^2 \)
Answer: To show that \( f(x, y) = \frac { x^2-y^2 }{ y^2+1 } \) is continuous at every \( (x, y) \in R^2 \), we need to confirm three conditions for continuity at an arbitrary point \( (a, b) \in R^2 \).
1. **\( f(a, b) \) must be defined:**
Substituting \( x=a \) and \( y=b \) into the function gives \( f(a, b) = \frac { a^2-b^2 }{ b^2+1 } \).
Since \( b^2 \ge 0 \), then \( b^2+1 \ge 1 \). The denominator \( b^2+1 \) is never zero.
Therefore, \( f(a, b) \) is defined for all \( (a, b) \in R^2 \).
2. **\( \lim _{(x, y) \rightarrow(a,b)} f(x, y) \) must exist:**
We need to evaluate \( \lim _{(x, y) \rightarrow(a,b)} \frac { x^2-y^2 }{ y^2+1 } \).
Since the denominator \( y^2+1 \) is not zero at \( (a,b) \), we can use direct substitution for polynomials and rational functions where the denominator is non-zero.
\( \lim _{(x, y) \rightarrow(a,b)} (x^2-y^2) = a^2-b^2 \)
\( \lim _{(x, y) \rightarrow(a,b)} (y^2+1) = b^2+1 \)
So, \( \lim _{(x, y) \rightarrow(a,b)} \frac { x^2-y^2 }{ y^2+1 } = \frac { \lim _{(x, y) \rightarrow(a,b)} (x^2-y^2) }{ \lim _{(x, y) \rightarrow(a,b)} (y^2+1) } = \frac { a^2-b^2 }{ b^2+1 } \).
The limit exists and is equal to \( \frac { a^2-b^2 }{ b^2+1 } \).
3. **\( \lim _{(x, y) \rightarrow(a,b)} f(x, y) = f(a, b) \):**
From condition 1, \( f(a, b) = \frac { a^2-b^2 }{ b^2+1 } \).
From condition 2, \( \lim _{(x, y) \rightarrow(a,b)} f(x, y) = \frac { a^2-b^2 }{ b^2+1 } \).
Since both are equal, the third condition is met.
Because all three conditions are satisfied for any arbitrary point \( (a, b) \in R^2 \), we conclude that \( f(x, y) \) is continuous at every point in \( R^2 \). This means there are no breaks or jumps in the function's graph.
In simple words: A function is continuous if it has no gaps or breaks. For this function, we checked three things at any point `(a,b)`: first, the function always has a value there; second, the limit as you get close to that point always exists; and third, the limit value is the same as the function's actual value at that point. Since all three are true for any `(a,b)`, the function is continuous everywhere.

🎯 Exam Tip: Remember the three conditions for continuity. For polynomial and rational functions, continuity can often be established by showing that the function is defined and that direct substitution works, especially when the denominator is never zero.

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 \( (0, 0) \)
Answer: To show that \( g(x, y) \) is continuous at \( (0, 0) \), we need to verify the three conditions for continuity.
1. **\( g(0, 0) \) must be defined:**
The problem statement explicitly gives \( g(0, 0) = 1 \). So, the function is defined at \( (0,0) \).
2. **\( \lim _{(x, y) \rightarrow(0,0)} g(x, y) \) must exist:**
We need to evaluate \( \lim _{(x, y) \rightarrow(0,0)} \frac { e^y \sin x }{ x } \).
We can separate the terms because their individual limits exist:
\( \lim _{(x, y) \rightarrow(0,0)} \frac { e^y \sin x }{ x } = \lim _{(x, y) \rightarrow(0,0)} e^y \cdot \lim _{(x, y) \rightarrow(0,0)} \frac { \sin x }{ x } \)
As \( (x, y) \rightarrow(0,0) \), it implies \( y \rightarrow 0 \) and \( x \rightarrow 0 \).
\( \lim _{y \rightarrow 0} e^y = e^0 = 1 \)
\( \lim _{x \rightarrow 0} \frac { \sin x }{ x } = 1 \) (This is a standard limit.)
So, \( \lim _{(x, y) \rightarrow(0,0)} g(x, y) = 1 \cdot 1 = 1 \).
The limit exists and is equal to 1.
3. **\( \lim _{(x, y) \rightarrow(0,0)} g(x, y) = g(0, 0) \):**
From condition 1, \( g(0, 0) = 1 \).
From condition 2, \( \lim _{(x, y) \rightarrow(0,0)} g(x, y) = 1 \).
Since both are equal to 1, the third condition is met.
As all three conditions are satisfied, we conclude that \( g(x, y) \) is continuous at \( (0, 0) \). Continuity at a point means the function's value matches the limit at that point.
In simple words: To check if a function is continuous at a point, we need to make sure three things are true: first, the function has a value at that point (which is given as 1 here); second, if you get very, very close to that point, the function's value also gets very close to a specific number (which is also 1 here); and third, these two numbers are the same. Since all three are true, the function is continuous at `(0,0)`.

🎯 Exam Tip: Pay careful attention to the definition of the function at the specific point, especially when it's defined separately (e.g., \( g(0,0) = 1 \)). This value must match the calculated limit for continuity.