How can I show $f_{n}(x) = sqrt{x^{2} + frac{1}{n}}$ is continuously differentiable on $x in ]-1,+1[$ for...
$begingroup$
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
49619
asked Dec 10 '18 at 2:56
josephjoseph
500111
$endgroup$
add a comment |
$begingroup$
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
49619
asked Dec 10 '18 at 2:56
josephjoseph
500111
$endgroup$
$begingroup$
you differentiated it using standard rules and the derivative made sense. I think that's enough
$endgroup$
– qbert
Dec 10 '18 at 3:05
add a comment |
$begingroup$
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
49619
asked Dec 10 '18 at 2:56
josephjoseph
500111
$endgroup$
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
49619
asked Dec 10 '18 at 2:56
josephjoseph
500111
edited Dec 10 '18 at 5:12
Neil hawking
49619
asked Dec 10 '18 at 2:56
josephjoseph
500111
edited Dec 10 '18 at 5:12
Neil hawking
49619
edited Dec 10 '18 at 5:12
Neil hawking
49619
edited Dec 10 '18 at 5:12
Neil hawking
49619
49619
asked Dec 10 '18 at 2:56
josephjoseph
500111
asked Dec 10 '18 at 2:56
josephjoseph
500111
asked Dec 10 '18 at 2:56
josephjoseph
500111
500111
$begingroup$
you differentiated it using standard rules and the derivative made sense. I think that's enough
$endgroup$
– qbert
Dec 10 '18 at 3:05
add a comment |
$begingroup$
you differentiated it using standard rules and the derivative made sense. I think that's enough
$endgroup$
– qbert
Dec 10 '18 at 3:05
$begingroup$
you differentiated it using standard rules and the derivative made sense. I think that's enough
$endgroup$
– qbert
Dec 10 '18 at 3:05
$begingroup$
you differentiated it using standard rules and the derivative made sense. I think that's enough
$endgroup$
– 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 GohSiong Thye Goh
101k1466117
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
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 GohSiong Thye Goh
101k1466117
$endgroup$
add a comment |
$begingroup$
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 GohSiong Thye Goh
101k1466117
$endgroup$
add a comment |
$begingroup$
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 GohSiong Thye Goh
101k1466117
$endgroup$
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 GohSiong Thye Goh
101k1466117
answered Dec 10 '18 at 3:56
Siong Thye GohSiong Thye Goh
101k1466117
answered Dec 10 '18 at 3:56
Siong Thye GohSiong Thye Goh
101k1466117
answered Dec 10 '18 at 3:56
Siong Thye GohSiong Thye Goh
101k1466117
101k1466117
add a comment |
add a comment |
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$begingroup$
you differentiated it using standard rules and the derivative made sense. I think that's enough
$endgroup$
– qbert
Dec 10 '18 at 3:05