I have functions $f,g,h:\mathbb{R}\rightarrow\mathbb{R}$ with: $$h(x)=cos(x)\cdot f(x)=e^x\cdot g(x)$$ If $h$ is continuous at $x_0$ and $g$ isn't continuous at $x_0$, then prove that $f$ isn't continuous at $x_0$. Any ideas on how to solve it, because I stack in this.
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$g(x)=e^{-x}\cos(x)f(x)$. If $f$ were continuous at $x_0$, what would happen ? – Gabriel Romon Nov 04 '17 at 08:50
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If $f$ is continuous at $x_0$ then $g(x)\cdot e^x=\cos (x)\cdot f(x)$. Since $e^x$ is non-zero so we can say $g(x)=\cos (x)\cdot f(x)/e^x$, which is continuous at $x_0$, a contradiction. – Sachchidanand Prasad Nov 04 '17 at 08:52
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Because $h$ is continuous at $x_0$, then $g(x)=h(x)e^{-x}$ is also continuous at $x_0$, so your claim is wrong.
szw1710
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split in two case:
1) $cos(x_0)\neq 0$ use $f(x)=\frac{e^xg(x)}{cos(x)}$ and use the discontinuity of g
2) $cos(x_0)=0$ use $f(x)=\frac{h(x)}{cos(x)}$ locally and use the discontinuity of $\frac{1}{cos(x)}$
polbos
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