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?
