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I'm trying to prove: Let $\lim_{x \rightarrow\ x_0 } f(x) = - L$ (for $L>0)$ and $\lim_{x \rightarrow\ x_0 } g(x) = \infty$ then $\lim_{x \rightarrow\ x_0 } f(x) \cdot g(x) = - \infty$.

I tried proving it using definition of limit by saying that $g(x) > K$ (for $K>0$) and tryed prove that $f(x) \cdot g(x) < K$ (but this is true only when $K<0$).

I don't know how to solve the issue that I have $K > 0$ but I need to finish with $K<0$.

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Note that, as $\lim_{x\to x_0} f(x)=-L$, pick a $L>\varepsilon>0$. There exists an $\delta_1$ so that $|x-x_0|<\delta_1$, then $|f(x)-(-L)|<\varepsilon$. That is, $-L-\varepsilon < f(x) < -L+\varepsilon <0$, so $f(x)<0$.

Now, fix $M>0$. We can find $\delta_2$ so that $|x-x_0|<\delta_2$, then $g(x)>M$.

To finish, if $M>0$, then call $\delta=\min\{\delta_1, \delta_2\}$, so both things above happen, that is, $f(x)<0$, and $g(x)>M$. Then you have: $$|x-x_0|<\delta \Rightarrow f(x)\cdot g(x) <(-L+\varepsilon)\cdot M$$

which is exactly $\lim_{x\to x_0} f(x)\cdot g(x)=-\infty$, because $\varepsilon$ and $L$ are fixed.

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    $\begingroup$ Just because $f(x) < 0$ and $g(x) > M$ does not imply that $f(x) g(x) < -M$ though. $\endgroup$ – guest Feb 12 '17 at 22:49
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    $\begingroup$ You are right, I already fixed that. Thanks for correcting my mistake. $\endgroup$ – A. Salguero-Alarcón Feb 12 '17 at 22:56

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