# 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?

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.