suppose $f : (a,b) \subset \mathbb{R} \to \mathbb{R}, x_0 \in (a,b)$ and $f(x) \to L$ as $x \to x_0$, with $L > 0$

1) I am trying to show there exists $\alpha > 0$ so that $f(x) > \frac{L}{2}$ when $x_0 - \alpha_1 < x < x_0 + \alpha_1$

2) prove $\frac{1}{f(x)} \to \frac{1}{L}$

my take(what I think it's helpful for this question):

$\displaystyle\lim_{x \to x_0^+} f(x) = L$ if for every $\epsilon>0$ and some $\alpha>0$ so that $|f(x)-L| < \epsilon$ when $x_0<x<x_0 + \alpha$.

$\displaystyle\lim_{x \to x_0^-} f(x) = L$ if for every $\epsilon>0$ and some $\alpha>0$ so that $|f(x)-L| < \epsilon$ when $x_0 - \alpha<x<x_0$

and if both $\displaystyle\lim_{x \to x_0^-} f(x) = L$, $\displaystyle\lim_{x \to x_0^+} f(x) = L$ then given $\epsilon >0$, I choose $\alpha$ to be the smaller $\alpha$ value s.t

$$|x-x_0| < \alpha_x, x < x_0$$ $$|x-x_0| < \alpha_y, x < x_0$$ $\alpha$ = min{$\alpha_x,\alpha_y$}

and these are all I could think of from the definition of Limit and functions.. Sorry my rough work is all over the place.. any help would be appreciated Thanks!


$(1)$ Given $\epsilon>0$ you know that there exists a $\delta>0$ such that $|x - x_0|< \delta \Rightarrow |f(x) - L|< \epsilon$. Let $\epsilon = L/2$ then we have a $\alpha$ such that;

$$|x-x_0|<\alpha \Rightarrow |f(x) - L| < \frac{L}{2}$$

Unraveling the above statements you have;

$$x_0 - \alpha<x< x_0+ \alpha \Rightarrow \frac{L}{2}<f(x)<\frac{3}{2} \cdot L \\$$

$(2)$ You want to show that given some $\epsilon >0$ there is some $\delta$ such that;

$$|x-x_0|< \delta \Rightarrow \left|\frac{1}{f(x)} - \frac{1}{L} \right|< \epsilon \iff \left|\frac{f(x)-L}{L \cdot f(x)}\right| = \frac{|f(x) - L|}{|L \cdot f(x)|}< \epsilon$$

Here we know that there exists $\delta$ such that;

$$|x-x_0|< \delta \Rightarrow |f(x)-L|< \epsilon \cdot |L \cdot f(x)| \Rightarrow \frac{|f(x) - L|}{|L \cdot f(x)|}< \frac{\epsilon}{|L \cdot f(x)|} \cdot |L \cdot f(x)| = \epsilon$$

Hence $\frac{1}{f(x)} \to \frac{1}{L}$ as desired.

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