Problems
a) Find $\lim_{x \rightarrow x_0^+}f(x)$ and $\lim_{x \rightarrow x_0^-}f(x)$ for $f(x) = \frac{x + |x|}{x}$, and $x_0 = 0$
b) Show that $\lim_{x \rightarrow -\infty}(1-\frac{1}{x^2})=1$
My attempt
a)
For the left hand side limit:
$$\forall \epsilon \gt 0 \, \,\exists \, \,\delta \, \,\ \,\text{s.t} \,\,|f(x) – L|<\epsilon \,\, \text{if} \, \, x_0 – \delta < x < x_0$$
$$ L = 0, x_0 = 0 $$
$$-\delta \lt x \lt 0\, \, , |\frac{x+ |x|}{x}| < \epsilon$$
For the right hand side limit:
$$\forall \epsilon \gt 0 \, \,\exists \, \,\delta \, \,\ \,\text{s.t} \,\,|f(x) – L|<\epsilon \,\, \text{if} \, \, x_0 < x < x_0 + \delta$$
$$L = 2, x_0 = 0$$
$$0 \lt x \lt \delta\, \, , |\frac{x+ |x|}{x} -2| < \epsilon$$
$$| \frac{2x}{x}| – 2= 0 \lt \epsilon $$
But how do I find expressions for delta for both the right-handside and left-handside limits?
b)
Per definition:
$\lim_{x \rightarrow -\infty} f(x) = L$ if $f(x)$ is defined on an interval $(-\infty, b)$, and for all $\epsilon \gt 0$ there is $\beta$ such that $|f(x) – L| < \epsilon$ if $x \lt \beta$.
Then:
$$|f(x) – L| = |1 – \frac{1}{x^2} – 1| = \frac{1}{x^2} \lt \epsilon \Rightarrow x \lt -\sqrt{\frac{1}{\epsilon}} $$
and thus $\beta = -\sqrt{\frac{1}{\epsilon}}$
Is this correct?
Best Answer
a) Since the function $f(x)$ is constant (in every side separately), so $\delta$ is arbitrary, for simplicity you can assume $\delta = \epsilon$.
b) This is correct, only you must add, $b < 0$, restriction.