Question

Indefinite limit problem:
Show that If an < 0 and lim an = 0, then lim 1/an = negative infinity.

Answer 1

Conditions

Indefinite limit problem:

Show that If an $< 0$ and $\lim \ an = 0$, then $\lim \ 1/an = \text{negative infinity}$.

Solution

Let's show that $\lim \frac{1}{\alpha_n} = -\infty$ by using a definition:

$$\lim x_n = -\infty \iff \forall \varepsilon > 0 \ \exists N = N(\varepsilon) \ \forall n \geq N \ x_n < -\varepsilon$$

Fix $\varepsilon > 0$

Let's find $N = N(\varepsilon): \forall n \geq N \ \frac{1}{\alpha_n} < -\varepsilon$

As we know, $\lim \alpha_n = 0$

$$\forall \varepsilon > 0 \ \exists N' = N'(\varepsilon) \ \forall n \geq N' \ |\alpha_n| < \varepsilon$$

As we know, $\alpha_n < 0 \ \forall n \in N$. So, we can remove the modulo in the next way:

$$\forall \varepsilon > 0 \ \exists N' = N'(\varepsilon) \ \forall n \geq N' - \alpha_n < \varepsilon$$ $$\alpha_n > -\varepsilon,$$ $$\frac{1}{\alpha_n} < \frac{1}{-\varepsilon} = -\varepsilon'$$

As this is true $\forall \varepsilon > 0$, so $\varepsilon' - \forall$. Here $N = N + 1$ (bigger than $N'$ for at least 1).