Compute the following limits.

• [math]\lim_{x \goesto 1} \frac{x^3-1}{x-1}[/math]
• [math]\lim_{x \goesto 2} \frac{x^2 - 5x +6}{x-2}[/math]
• [math]\lim_{x \goesto 3} \frac{x^2-5x+6}{x-3}[/math]
• [math]\lim_{x \goesto 1} \left( \frac{x^2}{x-1} - \frac{1}{x-1} \right)[/math]
• [math]\lim_{x \goesto 0} x^{\frac32}[/math]
• [math]\lim_{x \goesto 0} \frac{\sqrt{x+1}-1}x[/math]
• [math]\lim_{x \goesto 0} |x|[/math]
• [math]\lim_{h \goesto 0} \left( \frac{1-h^2}{h^2} + \frac{6h^2-1}{h^2} \right )[/math]
• [math]\lim_{x \goesto 0} \frac{(a+x)^3 + 2(a+x) - a^3 - 2a}x[/math]
• [math]\lim_{h \goesto 0} \frac{2(x+h)^2 - (x+h) - 2x^2 +x}h[/math].

For each of the following functions, find those numbers (if any) at which the function is not continuous.

• [math]x^3 + 3x - 1[/math]
• [math]f(x) = |x|[/math]
• [math]\frac{x^3+x+1}{x^2-x-2}[/math]
• [math]g(x) = \frac{x^3-3x-2}{x-2}[/math]
• [math]\sqrt{x+3}[/math]
• [math]h(x) = \frac{|x|}x[/math]
• [math]f(x) = \begin{cases} |x|, |x| \leq 1 \\ 2-x^2, |x| \gt 1 \end{cases}[/math]
• [math]F(x)= \begin{cases}1, \mbox{if $x$ is rational} \\ 0, \mbox{if $x$ is irrational}\end{cases}[/math]
• [math]f(x)= \begin{cases}x^2 +2x +1, \mbox{if $x \ne 1$} \\1, \mbox{if $x = 1$}\end{cases} [/math]

A function [math]f[/math] is said to have a removable discontinuity if it is not continuous at [math]a[/math], but can be assigned a value [math]f(a)[/math] [or possibly reassigned a new value [math]f(a)[/math]] such that it becomes continuous there.

• Locate the removable discontinuities in Problem Exercise.
• Show that the only discontinuities a rational function can have are either removable or infinite. That is, if [math]r(x)[/math] is a rational function that is not continuous at [math]a[/math], show that either [math]a[/math] is a removable discontinuity or [math]\lim_{x \goesto a} |r(x)| = + \infty[/math].

Using Theorem, prove that if [math]\lim_{x \goesto a} f(x) = b_1[/math] and [math]\lim_{x \goesto a} g(x) = b_2[/math], then

Show that

• [math]\lim_{x \goesto +\infty} f(x)=b[/math] if and only if [math]\lim_{t \goesto 0+} f \left( \frac1t \right) = b[/math].
• [math]\lim_{x \goesto -\infty} f(x)=b[/math] if and only if [math]\lim_{t \goesto 0-} f \left( \frac1t \right) = b[/math].

Using Problem Exercise, compute

• [math]\lim_{x \goesto \infty} \frac1{1+x}[/math]
• [math]\lim_{x \goesto \infty} \frac{3x+1}x[/math]
• [math]\lim_{t \goesto -\infty} \frac{4t^2 - 3t + 1}{t^2}[/math]
• [math]\lim_{t \goesto \infty} \frac{3t^3 + 7t^2 -2}{t^3+1}[/math].

True or false?

• If [math]\lim_{x \goesto a} f(x)=b[/math], then [math]\lim_{x \goesto a+} f(x)=b[/math] and [math]\lim_{x \goesto a-} f(x) = b[/math].
• If [math]\lim_{x \goesto a+} f(x)=b[/math] and [math]\lim_{x \goesto a-} f(x)=b[/math], then [math]\lim_{x \goesto a} f(x) = b[/math].

Define a function [math]f[/math] and draw its graph such that [math]\lim_{x \goesto 2+} f(x) = 2[/math] and [math]\lim_{x \goesto 2-} f(x) = 0[/math].

Compute

• [math]\lim_{x \goesto 2} \frac1{x^2 - 4x +4}[/math]
• [math]\lim_{x \goesto 2} \frac{x}{x^2 + 3x -10}[/math]
• [math]\lim_{x \goesto 3+} \frac{|x| - 3}{x-3}[/math]
• [math]\lim_{x \goesto 1} \frac{\sqrt{x-1}}{x-1}[/math]

Does the set of rational functions satisfy axioms through ?

(Hint: Be careful; note Exercise.)