Check whether the given function is continuous or not at the point (0, 0) | Query Point Official
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Check whether the given function is continuous or not at the point (0,
0).
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Solution:
We know that A function f of two variables is called continuous at the point `(x_0, y_0)` if
`f(x_0 , y_0)` is defined.
` lim_{(x,y) \to (x_0,y_0)} f(x,y)`exists.
` lim_{(x,y) \to (x_0,y_0)} f(x,y) = f(x_0,y_0)`
To determine, if f is continuous at (0,0), we need to compare ` lim_{(x,y) \to (x_0,y_0)} f(x,y) ` to `f(x_0,y_0)`
According to the definition of the function f, we find that
`f(0,0) = 1`
we now consider the limit ` lim_{(x,y) \to (x_0,y_0)} f(x,y)` of the function
`f(x,y) = \frac{x^2 + y^2 - x^2y^2}{x^2 + y^2}`
Let (x,y) approaches to the line x, where y=0
`f(x,y) = \frac{x^2 + 0^2 - x^20^2}{x^2 + 0^2}`
`f(x,y) = \frac{x^2 + 0 - 0}{x^2 + 0}`
`f(x,y) = \frac{x^2}{x^2}`
`f(x,y) = 1 - - - - (1)`
Let (x,y) approaches to the line y, where x=0
`f(x,y) = \frac{0^2 + y^2 - 0^2y^2}{0^2 + y^2}`
`f(x,y) = \frac{0 + y^2 - 0}{0 + y^2}`
`f(x,y) = \frac{y^2}{y^2}`
`f(x,y) = 1 - - - - - ( 2 )`
Let (x,y) approaches to the line, where y=x
`f(x,y) = \frac{x^2 + x^2 - x^2x^2}{x^2 + x^2}`
`f(x,y) = \frac{2x^2 - x^4}{2x^2}`
`f(x,y) = \frac{2x^2}{2x^2} - \frac{x^4}{2x^2}`
`f(x,y) = 1 - \frac{x^2}{2}`
By putting `f(x_0 , y_0) = (0,0)`
`f(x,y) = 1 - \frac{0^2}{2}`
`f(x,y) = 1 - - - - - ( 3 )`
From Equation (1),(2) and (3),
It is proved that the limit of the Above function exists and is equal to 1.
`f(x_0 , y_0) = 1`
`lim_{(x,y) \to (x_0,y_0} \right) f(x,y) = 1`
`lim_{(x,y) \to (x_0,y_0} \right) f(x,y) =f(x_0 , y_0) = 1`
Hence,
It is proved that the given function
is continuous or not at the point (0,0)
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Frequently Asked Questions (FAQs)
What does it mean for a function to be continuous at a point?
A function is said to be continuous at a point (x0, y0) if:
- The value f(x0, y0) is defined
- The limit lim(x,y)→(x₀,y₀) f(x,y) exists
- The limit equals the function value, i.e. lim(x,y)→(x₀,y₀) f(x,y) = f(x0, y0)
How do you check continuity of a function of two variables?
To check continuity of a function of two variables:
- Find the value of the function at the given point
- Evaluate the limit along different paths such as x = 0, y = 0, and y = x
- If all paths give the same limit and it equals the function value, the function is continuous
Why are different paths used to evaluate limits?
For functions of two variables, the limit may depend on the path taken. If the limit changes along different paths, then the limit does not exist and the function is not continuous at that point.
Is checking only two paths enough to prove continuity?
Checking two paths that give different results is sufficient to prove discontinuity. If multiple common paths give the same limit, it supports that the limit exists, especially in standard exam-level problems.
Is the given function continuous at (0,0)?
Yes. For the function:
f(x,y) = (x² + y² − x²y²) / (x² + y²)
The limit along the paths x = 0, y = 0, and y = x is 1, and the function value at (0,0) is also 1. Therefore, the function is continuous at (0,0).
Why is continuity important in calculus?
Continuity ensures smooth graphs and allows the use of differentiation and integration. It is essential in real-life applications such as motion, temperature distribution, and fluid flow.
Is this question important for exams?
Yes. Continuity of functions of two variables is a frequently asked question in:
- Engineering Mathematics
- BSc / BS Mathematics
- University semester and competitive exams
Such questions usually carry 5–8 marks.
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