Solution. From the left:

From the right:

Both columns close in on \(4\), so \[\lim _{x\to 2^-}x^2=4,\qquad \lim _{x\to 2^+}x^2=4,\qquad \text {hence}\qquad \lim _{x\to 2}x^2=4.\]

Here the limit equals \(f(2)=4\), which is what makes this function continuous at \(x=2\). That is a convenience, not a rule.

Solution. (a) Substitute directly: \[3(2)^2+2-5=12+2-5=9.\]

(b) The denominator at \(x=1\) is \(3\neq 0\), so substitution is safe: \[\frac {1-2+7}{1+2}=\frac {6}{3}=2.\]

(c) Substituting gives \(\frac {4+4-8}{2-2}=\frac {0}{0}\), which is meaningless. But \(\frac {0}{0}\) arising means the numerator and denominator share a factor of \((x-2)\). Factorise: \[\frac {x^2+2x-8}{x-2}=\frac {(x-2)(x+4)}{x-2}=x+4\qquad \text {for } x\neq 2.\] Cancelling is legitimate precisely because \(x\) never equals \(2\) in a limit — it only approaches it. So \[\lim _{x\to 2}\frac {x^2+2x-8}{x-2}=\lim _{x\to 2}(x+4)=6.\]

Note that \(f(2)\) itself does not exist here, yet the limit does. This is exactly the situation the definition was written to handle, and it is the situation a derivative will present in a moment.

Solution. (a) Approaching from each side uses a different rule, so the two one-sided limits must be worked out separately.

From the left, \(x\) is below \(3\) and the first rule applies: \[\lim _{x\to 3^-}f(x)=\lim _{x\to 3^-}\left (x^2-1\right )=9-1=8.\] From the right, \(x\) is above \(3\) and the second rule applies: \[\lim _{x\to 3^+}f(x)=\lim _{x\to 3^+}(2x+1)=6+1=7.\] The two disagree, so \[\therefore \quad \lim _{x\to 3}f(x)\ \text {does not exist}.\] The graph jumps at \(x=3\), and a function heading for two different values at once is heading for neither.

(b) The value at \(x=1\) is read off the second line directly: \[g(1)=10.\] The limit, however, never uses the value at \(x=1\) — only the values nearby, which all come from the first line: \[\lim _{x\to 1}g(x)=\lim _{x\to 1}\left (x^2+3\right )=1+3=4.\]

\[\therefore \quad g(1)=10\quad \text {but}\quad \lim _{x\to 1}g(x)=4.\]

Solution. (a) The highest power below is \(x\), so divide throughout by \(x\): \[\frac {3x+2}{4x-1}=\frac {3+\frac {2}{x}}{4-\frac {1}{x}}.\] As \(x\to \infty \) both \(\frac {2}{x}\) and \(\frac {1}{x}\) go to \(0\): \[\lim _{x\to \infty }\frac {3x+2}{4x-1}=\frac {3+0}{4-0}=\frac {3}{4}.\]

(b) Divide throughout by \(x^2\): \[\frac {7x^2-x+4}{2-3x^2}=\frac {7-\frac {1}{x}+\frac {4}{x^2}}{\frac {2}{x^2}-3} \longrightarrow \frac {7-0+0}{0-3}=-\frac {7}{3}.\]

(c) The denominator has the higher degree, so divide by \(x^3\): \[\frac {4x-1}{x^3+2}=\frac {\frac {4}{x^2}-\frac {1}{x^3}}{1+\frac {2}{x^3}} \longrightarrow \frac {0-0}{1+0}=0.\] The bottom outgrows the top, so the fraction is squeezed to zero.

(d) Now the numerator has the higher degree. Dividing by \(x\): \[\frac {x^3+2x}{5x-1}=\frac {x^2+2}{5-\frac {1}{x}},\] and the top grows without bound while the bottom settles at \(5\). So \[\lim _{x\to \infty }\frac {x^3+2x}{5x-1}\ \text {does not exist; the function tends to } \infty .\]