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[Hilmaam]

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To show that a function F is continuous on the real numbers (R), we need to show that for any point x in R, and for any sequence of points {x_n} that converges to x, the sequence {F(x_n)} also converges to F(x).

In this case, the function F is defined as f(x) = x^3/(1+x^2). We can show that this function is continuous on R as follows:

Let x be any point in R, and let {x_n} be a sequence of points in R that converges to x. We need to show that the sequence {F(x_n)} = {x_n^3/(1+x_n^2)} also converges to F(x) = x^3/(1+x^2).

To do this, we can use the fact that the sum, difference, product, and quotient of two continuous functions are also continuous. We know that the functions x^3 and 1+x^2 are both continuous on R, so their quotient, F(x) = x^3/(1+x^2), is also continuous on R.

Therefore, F is continuous on R.

As for uniform continuity, a function is uniformly continuous on a set if, given any positive number ε, there exists a positive number δ such that for all points x and y in the set, if the distance between x and y is less than δ, then the distance between f(x) and f(y) is also less than ε.

In this case, we can show that the function F is not uniformly continuous on R. To do this, we can consider the sequence of points {x_n} = {(-1)^n/n}, which oscillates between positive and negative values and approaches 0 as n approaches infinity.

For this sequence, we have that F(x_n) = x_n^3/(1+x_n^2) = (-1)^n/n^3/(1+1/n^2) = (-1)^n/n, which also oscillates between positive and negative values but does not approach any particular value as n approaches infinity.

Therefore, F is not uniformly continuous on R, because for any positive ε, we can find points x and y in R such that the distance between x and y is arbitrarily small (less than δ for any positive δ), but the distance between F(x) and F(y) is not less than ε.

 

[Hilmaam]

To show that a function F is continuous on the real numbers (R), we need to show that for any point x in R, and for any sequence of points {x_n} that converges to x, the sequence {F(x_n)} also converges to F(x).

In this case, the function F is defined as f(x) = x^3/(1+x^2). We can show that this function is continuous on R as follows:

Let x be any point in R, and let {x_n} be a sequence of points in R that converges to x. We need to show that the sequence {F(x_n)} = {x_n^3/(1+x_n^2)} also converges to F(x) = x^3/(1+x^2).

To do this, we can use the fact that the sum, difference, product, and quotient of two continuous functions are also continuous. We know that the functions x^3 and 1+x^2 are both continuous on R, so their quotient, F(x) = x^3/(1+x^2), is also continuous on R.

Therefore, F is continuous on R.

As for uniform continuity, a function is uniformly continuous on a set if, given any positive number ε, there exists a positive number δ such that for all points x and y in the set, if the distance between x and y is less than δ, then the distance between f(x) and f(y) is also less than ε.

In this case, we can show that the function F is not uniformly continuous on R. To do this, we can consider the sequence of points {x_n} = {(-1)^n/n}, which oscillates between positive and negative values and approaches 0 as n approaches infinity.

For this sequence, we have that F(x_n) = x_n^3/(1+x_n^2) = (-1)^n/n^3/(1+1/n^2) = (-1)^n/n, which also oscillates between positive and negative values but does not approach any particular value as n approaches infinity.

Therefore, F is not uniformly continuous on R, because for any positive ε, we can find points x and y in R such that the distance between x and y is arbitrarily small (less than δ for any positive δ), but the distance between F(x) and F(y) is not less than ε.

To show that a function is continuous at a point, we need to show that the function's values get arbitrarily close to the value at the point as we get arbitrarily close to the point with the input of the function.

For the function f defined as f(x) = x^3/(1+x^2), we can show that it is continuous on the entire real number line, R, by showing that it is continuous at every point on R. Let's consider an arbitrary point c on R. We need to show that for any small positive number ε, there exists a small positive number δ such that for all x, if |x - c| < δ, then |f(x) - f(c)| < ε.

We can do this by finding an expression for |f(x) - f(c)| in terms of |x - c|, and then showing that this expression is always less than ε for a suitable choice of δ.

We have:

|f(x) - f(c)| = |x^3/(1+x^2) - c^3/(1+c^2)|

= |x^3 - c^3| / |1+x^2| + |c^3| / |1+x^2| * |1/(1+c^2) - 1/(1+x^2)|

Now, we can bound the second term on the right-hand side as follows:

|c^3| / |1+x^2| * |1/(1+c^2) - 1/(1+x^2)| <= |c^3| / |1+x^2| * |1/(1+c^2)|

Since |1+x^2| is always positive and |1/(1+c^2)| is always positive and less than or equal to 1, this term is always less than or equal to |c^3|.

Therefore, we have:

|f(x) - f(c)| <= |x^3 - c^3| / |1+x^2| + |c^3|

Now, we can bound the first term on the right-hand side as follows:

|x^3 - c^3| / |1+x^2| <= |x - c| * (|x^2| + |c^2|) / |1+x^2|

Since |1+x^2| is always positive, this term is always less than or equal to |x - c| * (|x^2| + |c^2|).

Therefore, we have:

|f(x) - f(c)| <= |x - c| * (|x^2| + |c^2|) + |c^3|

Now, we can choose δ to be a small positive number such that δ * (|c^2| + |c^2|) < ε. Then, for any x such that |x - c| < δ, we have:

|f(x) - f(c)| <= |x - c| * (|x^2| + |c^2|) + |c^3| < δ * (|c^2| + |c^2|) + |c^3| < ε

This shows that f is continuous at the point c. Since c was an arbitrary point on R, this shows

 

[Hilmaam]

To show that a function is continuous at a point, we need to show that the function's values get arbitrarily close to the value at the point as we get arbitrarily close to the point with the input of the function.

For the function f defined as f(x) = x^3/(1+x^2), we can show that it is continuous on the entire real number line, R, by showing that it is continuous at every point on R. Let's consider an arbitrary point c on R. We need to show that for any small positive number ε, there exists a small positive number δ such that for all x, if |x - c| < δ, then |f(x) - f(c)| < ε.

We can do this by finding an expression for |f(x) - f(c)| in terms of |x - c|, and then showing that this expression is always less than ε for a suitable choice of δ.

We have:

|f(x) - f(c)| = |x^3/(1+x^2) - c^3/(1+c^2)|

= |x^3 - c^3| / |1+x^2| + |c^3| / |1+x^2| * |1/(1+c^2) - 1/(1+x^2)|

Now, we can bound the second term on the right-hand side as follows:

|c^3| / |1+x^2| * |1/(1+c^2) - 1/(1+x^2)| <= |c^3| / |1+x^2| * |1/(1+c^2)|

Since |1+x^2| is always positive and |1/(1+c^2)| is always positive and less than or equal to 1, this term is always less than or equal to |c^3|.

Therefore, we have:

|f(x) - f(c)| <= |x^3 - c^3| / |1+x^2| + |c^3|

Now, we can bound the first term on the right-hand side as follows:

|x^3 - c^3| / |1+x^2| <= |x - c| * (|x^2| + |c^2|) / |1+x^2|

Since |1+x^2| is always positive, this term is always less than or equal to |x - c| * (|x^2| + |c^2|).

Therefore, we have:

|f(x) - f(c)| <= |x - c| * (|x^2| + |c^2|) + |c^3|

Now, we can choose δ to be a small positive number such that δ * (|c^2| + |c^2|) < ε. Then, for any x such that |x - c| < δ, we have:

|f(x) - f(c)| <= |x - c| * (|x^2| + |c^2|) + |c^3| < δ * (|c^2| + |c^2|) + |c^3| < ε

This shows that f is continuous at the point c. Since c was an arbitrary point on R, this shows

Ran question twice through tool. have no idea if it’s even close to right , been a while since I was in school. Link to the tool if you wanna try it https://chat.openai.com/chat

 

[bidenkulaha]

Ran question twice through tool. have no idea if it’s even close to right , been a while since I was in school. Link to the tool if you wanna try it https://chat.openai.com/chat

It’s defo uniformly continuous using the Lipschitz definition lol. AI needs some work for analysis