# Show that if $\lim_{x\rightarrow x_0} f(x)=L>0$ then $\lim_{x\rightarrow x_0} 1/f(x)=1/L>0$

Show that if $\lim_{x\rightarrow x_0} f(x)=L>0$ then $\lim_{x\rightarrow x_0} 1/f(x)=1/L>0$

My proof is right now using the fact that $\delta = \epsilon/L$. I'm not sure whether this is correct at all.

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Given $\epsilon>0$ you want to find $\delta>0$ such that $|x-x_0|<\delta$ implies $\left|\frac1{f(x)}-\frac1L\right|<\epsilon$. Observer that $\frac1{f(x)}-\frac1L = \frac{f(x)-L}{f(x)L}$. You can make sure that the numerator is small because $f(x)\to L$. But how can you prevent the denominator from getting small (which migh tmake the fraction big)? By a suitable choice of $\delta$ you can make sure that $f(x)>\frac L2$ (namely, that $|f(x)-L|<\frac L2$). This makes $\left|\frac1{f(x)}-\frac1L\right|<\frac2{L^2}\cdot |f(x)-L|$, hence "manageable". You must make your $delta$ small enough to fulfill two conditions, that is let $\delta=\min\{\delta_1,\delta_2\}$ where $\delta_1$ is chosen to warrant $f(x)>\frac L2$ and $\delta_2$ is chosen to warrant $|f(x)-L|<\frac{L^2}2\cdot\epsilon$.