Solution. (a). \(f(2)\) is undefined, since both numerator and denominator vanish, so the first condition already fails. The limit however exists: for \(x\neq 2\), \[\frac {x^{2}-4}{x-2}=\frac {(x-2)(x+2)}{x-2}=x+2\longrightarrow 4 .\] A limit exists but no value is attached to it, so the discontinuity is removable. Defining \(f(2)=4\) makes the function continuous.

(b). \(g(3)\) is undefined, and the limit does not exist either: as \(x\to 3\) from either side, \((x-3)^{2}\) is a small positive number and its reciprocal grows without bound. This is an infinite discontinuity.

(c). For \(x>0\), \(\left |x\right |=x\) so \(h(x)=1\); for \(x<0\), \(\left |x\right |=-x\) so \(h(x)=-1\). Hence \[\lim _{x\to 0^{-}}h(x)=-1,\qquad \lim _{x\to 0^{+}}h(x)=1 .\] Both one-sided limits exist and differ, so this is a jump discontinuity of size \(2\). (The function is also undefined at \(0\), but that is not what makes it unrepairable — no value at \(0\) could bridge a gap between \(-1\) and \(1\).)

Solution. Each branch is a polynomial and so continuous on its own piece; only the join needs attention. The three conditions require the value and the two one-sided limits to agree.

Since \(x\leq 2\) uses the first branch, \(f(2)=2k+1\), and \[\lim _{x\to 2^{-}}f(x)=2k+1,\qquad \lim _{x\to 2^{+}}f(x)=2^{2}-1=3 .\] The left limit already matches the value, so continuity needs only \[2k+1=3 \implies k=1 .\] With \(k=1\) both sides give \(3\) and the function is continuous. For any other \(k\) there is a jump at \(x=2\) of size \(\left |3-(2k+1)\right |\).

Solution. (a). Substitution gives \(\frac {0}{0}\), so factor. Both are differences of powers with \(x-2\) as a factor: \[x^{2}-4=(x-2)(x+2),\quad x^{3}-8=(x-2)\left (x^{2}+2x+4\right ).\] Cancelling the common factor, \[\lim _{x\to 2}\frac {x+2}{x^{2}+2x+4}=\frac {4}{4+4+4}=\frac {1}{3}.\]

(b). An \(\frac {\infty }{\infty }\) form. Divide above and below by \(x^{3}\): \[\lim _{x\to \infty }\frac {3+\frac {5}{x^{3}}}{\frac {5}{x^{3}}-\frac {1}{x}-3} =\frac {3+0}{0-0-3}=-1 .\]