How can I show $f_{n}(x) = sqrt{x^{2} + frac{1}{n}}$ is continuously differentiable on $x in ]-1,+1[$ for...
I want to show $f_{n}(x) = sqrt{x^{2} + frac{1}{n}}$ is continuously differentiable for each $x in ]-1,+1[$ and $n in mathbb{N}$?
I have
$$f'_{n}(x) = frac{x}{sqrt{x^{2} + frac{1}{n}}}.$$
It is obviously continuous because the denominator is always positive. But is there a more rigorous way to show this? Or is this sufficient? Since $n in mathbb{N}$, the $ frac{1}{n}$ is always positive. Also $x^{2}$ is the square of a number, so it's always positive. I just want to show continuity though.
sequences-and-series continuity
edited Dec 10 '18 at 5:12
Neil hawking
48819
asked Dec 10 '18 at 2:56
joseph
4329
add a comment |
I want to show $f_{n}(x) = sqrt{x^{2} + frac{1}{n}}$ is continuously differentiable for each $x in ]-1,+1[$ and $n in mathbb{N}$?
I have
$$f'_{n}(x) = frac{x}{sqrt{x^{2} + frac{1}{n}}}.$$
It is obviously continuous because the denominator is always positive. But is there a more rigorous way to show this? Or is this sufficient? Since $n in mathbb{N}$, the $ frac{1}{n}$ is always positive. Also $x^{2}$ is the square of a number, so it's always positive. I just want to show continuity though.
sequences-and-series continuity
edited Dec 10 '18 at 5:12
Neil hawking
48819
asked Dec 10 '18 at 2:56
joseph
4329
you differentiated it using standard rules and the derivative made sense. I think that's enough
– qbert
Dec 10 '18 at 3:05
add a comment |
I want to show $f_{n}(x) = sqrt{x^{2} + frac{1}{n}}$ is continuously differentiable for each $x in ]-1,+1[$ and $n in mathbb{N}$?
I have
$$f'_{n}(x) = frac{x}{sqrt{x^{2} + frac{1}{n}}}.$$
It is obviously continuous because the denominator is always positive. But is there a more rigorous way to show this? Or is this sufficient? Since $n in mathbb{N}$, the $ frac{1}{n}$ is always positive. Also $x^{2}$ is the square of a number, so it's always positive. I just want to show continuity though.
sequences-and-series continuity
edited Dec 10 '18 at 5:12
Neil hawking
48819
asked Dec 10 '18 at 2:56
joseph
4329
I want to show $f_{n}(x) = sqrt{x^{2} + frac{1}{n}}$ is continuously differentiable for each $x in ]-1,+1[$ and $n in mathbb{N}$?
I have
$$f'_{n}(x) = frac{x}{sqrt{x^{2} + frac{1}{n}}}.$$
It is obviously continuous because the denominator is always positive. But is there a more rigorous way to show this? Or is this sufficient? Since $n in mathbb{N}$, the $ frac{1}{n}$ is always positive. Also $x^{2}$ is the square of a number, so it's always positive. I just want to show continuity though.
sequences-and-series continuity
sequences-and-series continuity
edited Dec 10 '18 at 5:12
Neil hawking
48819
asked Dec 10 '18 at 2:56
joseph
4329
edited Dec 10 '18 at 5:12
Neil hawking
48819
asked Dec 10 '18 at 2:56
joseph
4329
edited Dec 10 '18 at 5:12
Neil hawking
48819
edited Dec 10 '18 at 5:12
Neil hawking
48819
edited Dec 10 '18 at 5:12
Neil hawking
48819
48819
asked Dec 10 '18 at 2:56
joseph
4329
asked Dec 10 '18 at 2:56
joseph
4329
asked Dec 10 '18 at 2:56
joseph
4329
4329
you differentiated it using standard rules and the derivative made sense. I think that's enough
– qbert
Dec 10 '18 at 3:05
add a comment |
you differentiated it using standard rules and the derivative made sense. I think that's enough
– qbert
Dec 10 '18 at 3:05
you differentiated it using standard rules and the derivative made sense. I think that's enough
– qbert
Dec 10 '18 at 3:05
you differentiated it using standard rules and the derivative made sense. I think that's enough
– qbert
Dec 10 '18 at 3:05
add a comment |
1 Answer
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I believe that's enough.
It should be clear that $x^2+frac1n$, a polynomial, is continuous. Composition of continuos function is continuous, Hence $sqrt{x^2+frac1n}$ is continuous.
A continuous function divided by a non-zero continuous function gives us another continuous function, hence your function is continuous.
answered Dec 10 '18 at 3:56
Siong Thye Goh
99.4k1464117
add a comment |
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1 Answer
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1 Answer
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I believe that's enough.
It should be clear that $x^2+frac1n$, a polynomial, is continuous. Composition of continuos function is continuous, Hence $sqrt{x^2+frac1n}$ is continuous.
A continuous function divided by a non-zero continuous function gives us another continuous function, hence your function is continuous.
answered Dec 10 '18 at 3:56
Siong Thye Goh
99.4k1464117
add a comment |
I believe that's enough.
It should be clear that $x^2+frac1n$, a polynomial, is continuous. Composition of continuos function is continuous, Hence $sqrt{x^2+frac1n}$ is continuous.
A continuous function divided by a non-zero continuous function gives us another continuous function, hence your function is continuous.
answered Dec 10 '18 at 3:56
Siong Thye Goh
99.4k1464117
add a comment |
I believe that's enough.
It should be clear that $x^2+frac1n$, a polynomial, is continuous. Composition of continuos function is continuous, Hence $sqrt{x^2+frac1n}$ is continuous.
A continuous function divided by a non-zero continuous function gives us another continuous function, hence your function is continuous.
answered Dec 10 '18 at 3:56
Siong Thye Goh
99.4k1464117
I believe that's enough.
It should be clear that $x^2+frac1n$, a polynomial, is continuous. Composition of continuos function is continuous, Hence $sqrt{x^2+frac1n}$ is continuous.
A continuous function divided by a non-zero continuous function gives us another continuous function, hence your function is continuous.
answered Dec 10 '18 at 3:56
Siong Thye Goh
99.4k1464117
answered Dec 10 '18 at 3:56
Siong Thye Goh
99.4k1464117
answered Dec 10 '18 at 3:56
Siong Thye Goh
99.4k1464117
answered Dec 10 '18 at 3:56
Siong Thye Goh
99.4k1464117
99.4k1464117
add a comment |
add a comment |
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you differentiated it using standard rules and the derivative made sense. I think that's enough
– qbert
Dec 10 '18 at 3:05