This presentation discusses limits and continuity of functions of several variables. It defines limits of multivariable functions and provides examples of limits that do and do not exist as the variables approach certain points. It also defines continuity as a function being continuous at a point if the limit of the function as the variables approach that point equals the function value at that point. Examples show functions that are and aren't continuous based on whether their limits exist or not.

1. PRESENTATION FOR
CALCULUS
TOPIC :- LIMIT AND CONTINUTY OF
THE FUNCTION OF SEVERAL
VARIABLES

2. LIMIT OF FUNCTION
• We say that a function f(x , y) approaches the limit
L as (x , y) approaches (xₒ , yₒ) ,and written as,
• Lim f(x , y) = L
(x , y)→(x₀ , y₀)
• For every ε>0,there exists δ>0 such that, |f(x,y)-
L|<ε,whenever
d((x , y),(x₀ , y₀))< δ

3. EXAMPLES
• 1.]
lim [(xy)/(x²+y²)]
(x , y) →(0 , 0)
→ lim
(x , y) →(0 , 0) [(xy)/(x²+y²)]
y=0
=0

4. → lim
(x , y) →(0 , 0) [(xy)/(x²+y²)]
x=0
=0
→ lim
(x , y) →(0 , 0) [(xy)/(x²+y²)]
y=x
= x²/2x²
=1/2

5. → lim (*D.N.E.=Does not exist)
(x , y) →(0 , 0) [(xy)/(x²+y²)]
y=-x
= -x²/2x²
=-1/2
Hence ,
lim
(x , y) →(0 , 0) [(xy)/(x²+y²)] ≠
y=x
lim
(x , y) →(0 , 0) [(xy)/(x²+y²)]
y=-x
So limit D.N.E.

6. • 2.]
lim [(x²y)/(x⁴+y²)]
(x , y) →(0 , 0)
→y=mx²
= lim
(x , y) →(0 , 0) [x²(mx²)/x⁴+m² x⁴]
y=mx²
= m/1+m²
• Which is independent from x & y.
• So the limit D.N.E.

7. • 3.]
lim [(x²-xy)/(√x-√y)]
(x , y) →(0 , 0)
= lim [x(x-y)/(√x-√y)]
(x , y) →(0 , 0)
= lim [x(√x-√y) (√x+√y)]/(√x-√y)
(x , y) →(0 , 0)
= lim x(√x+√y)
(x , y) →(0 , 0)
=0
• So the limit exists and its value is 0.

8. • 4.]
lim [(x³-y³)/(x²+y²)]
(x , y) →(0 , 0)
→ Taking y = mxⁿ
= lim
(x , y) →(0 , 0) (x³-m³x³ⁿ)/(x²+m²x²ⁿ)
y=mxⁿ
= lim x[(1-m³x³ⁿ̄̄̄̄ˉ³)/(1+m²x²ⁿ̄̄̄̄ˉ²)]
(x , y) →(0 , 0)

9. = 0
• So the limit exists and its value is 0.
5.] Find :- lim (x⁴ + 4x³y - 5xy²)
(x , y) →(5 , -2)
→ lim (x⁴ + 4x³y - 5xy²)
(x , y) →(5 , -2)
= (5)⁴ + 4(5)(-2)³ - 5(5)(-2)²
=-475

10. CONTINUTY OF A FUNCTION
• The function u=f(x , y) is said to be continuous at
the point (x₀ , y₀), if for all points (x , y) near (x₀ , y₀)
the value of f(x , y) differs , but little , from value
f(x₀ , y₀). In the other words if f has the domain R
and Q = (x₀ , y₀) is a point of R, then f is continuous
at Q if for every ε>0 there exists a δ>0 such that
|f(P) – f(Q)| = |f(x , y) – f(x₀ , y₀)| < ε

11. • For all P = (x , y) in R such for which
d((x , y),(x₀ , y₀)) = √[(x-x₀)² + (y-y₀)²]< δ
• Let f be a function of two variables x and y & (x₀ ,
y₀) be a point of R² , then f is continuous at (x₀ , y₀)
if,
limit f(x , y) = f (x₀ , y₀)
(x , y)→ (x₀ , y₀)

12. EXAMPLES
• 1.] f(x , y)= 2xy/x²+y² , if (x , y)≠(0 , 0)
= 0 , if (x , y)=(0 , 0)
→ lim
(x , y) →(0 , 0) [2x(mx)/x²+y²]
y=mx
=2m/(1+m²)
Which only depends on m.
So the limit D.N.E.
So the function can’t be continuous at (0 , 0).

13. • 2.]
f(x , y) = x²-y²/x²+y² , if (x , y) ≠ (0 , 0)
= 0 , if (x , y) = (0 , 0).
→ lim
(x , y) →(0 , 0) [x²-m²x²/x²+m²x²]
y=mx
=(1-m²)/(1+m²)
Which only depends on m.
So the limit D.N.E.
So the function can’t be continuous at (0 , 0).

14. THANK YOU
• PRESENTED BY :-
• BHAGYESH PATEL(22)
• UTKARSH GANDHI(23)
• KUNAL PATIL(24)
• GUIDED BY :-
• SWAGAT SIR