Explain the continuity of the function $f$ defined by $f(x) = \lim_{n\to \infty} \frac{e^x-x^n\sin(nx)}{1+x^n} (0\leq x\leq \pi/\sqrt2)$ at $x = 1$.

Asked Feb 21 '24 at 08:54

Active Feb 21 '24 at 12:06

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Explain for continuity the function $f$ defined by $f(x) = \lim_{n\to \infty} \frac{e^x-x^n\sin(nx)}{1+x^n} (0\leq x\leq \pi/ \sqrt2)$ at $x = 1$. Explain why the function $f$ does not vanish anywhere in $[0, \pi/\sqrt2]$ although $f(0)f(\pi/\sqrt2) <0$.

My attemt

If $0 \leq x < 1$ then $x^n$ $\to$ 0 therefore $f(x) = \lim_{n\to \infty} \frac{e^x-x^n\sin(nx)}{1+x^n} = e^x$.

If $ x > 1$ then $x^n \to ∞$ therefore $f(x) = \lim_{n\to \infty} \frac{e^x-x^n\sin(nx)}{1+x^n} = -\sin(nx)$

I don't know how to proceed further please help me

edited Feb 21 '24 at 12:06

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asked Feb 21 '24 at 08:54

Newuser1155953

$\lim_{n\to \infty} \frac{e^x-x^n\sin(nx)}{1+x^n} = -\sin(nx)$ makes no sense. The limit must not depend on $n$. – Anne Bauval Feb 21 '24 at 09:26
What do you mean by "Explain for continuity"? – Anne Bauval Feb 21 '24 at 09:38
3

Anyway, the whole exercise (btw, where does it come from?) makes no sense because $f(x)$ is not well-defined if $1<x<\pi$. https://math.stackexchange.com/questions/1857512 – Anne Bauval Feb 21 '24 at 09:45
@AnneBauval sorry it's $\sqrt2$ – Newuser1155953 Feb 21 '24 at 11:42
1

Could it be $\sin^n(x)$ instead of $\sin(nx)$? Maybe incorrectly copied from the source? – Ingix Feb 21 '24 at 13:26

Explain the continuity of the function $f$ defined by $f(x) = \lim_{n\to \infty} \frac{e^x-x^n\sin(nx)}{1+x^n} (0\leq x\leq \pi/\sqrt2)$ at $x = 1$.

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