I'm reading a solution to the following exercise:
"Assume that $\lim_{x\to c}f\left(x\right)=L$, where $L\ne0$, and assume $\lim_{x\to c}g\left(x\right)=0.$ Show that $\lim_{x\to c}\left|\frac{f\left(x\right)}{g\left(x\right)}\right|=\infty.$"
And at some point in the proof the following step appears:
"Choose $\delta_1$ so that $0<\left|x-c\right|<\delta _1$ implies $\left|f\left(x\right)-L\right|<\frac{|L|}{2}$. Then we have $\left|f\left(x\right)\right|\ge\frac{\left|L\right|}{2}$."
It's precisely the implication in bold that I'm struggling to understand. How does the writer go from $\left|f\left(x\right)-L\right|<\frac{\left|L\right|}{2}$ to $\left|f\left(x\right)\right|\ge\frac{\left|L\right|}{2}$?
I'm probably failing to see something that may be very clear, but I've been attempting unsuccessfully to reach the conclusion algebraically long enough, and can't quite see why it is true either!
Here is the rest of the solution, if necessary.
"Let $M>0\ $ be arbitrary. [...]. Because $\lim_{x\to c}g\left(x\right)=0$, we can choose $\delta_2$ such that $\left|g\left(x\right)\right|<\frac{\left|L\right|}{2M}\ $provided $0<\left|x-c\right|<\delta_2$.
Let $\delta=\min\left\{\delta_1,\delta_2\right\}.\ $ Then we have
$\left|\frac{f\left(x\right)}{g\left(x\right)}\right|\ge\left|\frac{\frac{\left|L\right|}{2}}{\frac{\left|L\right|}{2M}}\right|=M$ provided $0<\left|x-c\right|<\delta$, as desired."
Look at this picture:
If $|f(x) - L| < \frac {|L|}2$ then $f(x)$ must be in that orange circle.
But everything within the orange circle is within $\frac{|L|}2$ of $L$ which means it is more than $\frac{|L|}2$ away from the origin.
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Now to do it without pictures we use the triangle inequality.
Confession, I always mess this up.
But using the picture as a guideline I get
$|L-f(x)| + |f(x)-0| \ge |L - 0|$ and .... then it falls into place
$|f(x) - L| + |f(x)| \ge |L|$ so
$|f(x)| \ge |L| - |f(x) - L| > |L| - \frac {|L|}2 = \frac {|L|}2$.