Solution.

(a)

\(\lim \limits _{h\longrightarrow 0}\frac {\sqrt {x+h}-\sqrt {x}}{h}\). Rationalise the numerator: \begin {align*} \lim _{h\longrightarrow 0}\frac {(\sqrt {x+h}-\sqrt {x})(\sqrt {x+h}+\sqrt {x})}{h(\sqrt {x+h}+\sqrt {x})} & =\lim _{h\longrightarrow 0}\frac {(\sqrt {x+h})^2-(\sqrt {x})^2}{h(\sqrt {x+h}+\sqrt {x})}\\ & =\lim _{h\longrightarrow 0}\frac {x+h-x}{h(\sqrt {x+h}+\sqrt {x})}\\ & =\lim _{h\longrightarrow 0}\frac {1}{\sqrt {x+h}+\sqrt {x}}\\ & = \frac {1}{\sqrt {x}+\sqrt {x}}\\ & = \frac {1}{2\sqrt {x}}\\ \end {align*}

(b)

\(\lim \limits _{x\longrightarrow 2}\frac {3x-1}{x}\). Substitution is legitimate here — the denominator is not zero at \(x=2\) — and gives \(\frac {3(2)-1}{2}=\frac {5}{2}\).

(c)

\(\lim \limits _{x\longrightarrow 2}\frac {4-x^2}{3-\sqrt {x^2+5}}\). Substitution gives \(\frac {0}{0}\), and the root sits in the denominator, so rationalise there: \begin {align*} \lim _{x\to 2}\frac {4-x^{2}}{3-\sqrt {x^{2}+5}} &=\lim _{x\to 2}\frac {\left (4-x^{2}\right )\left (3+\sqrt {x^{2}+5}\right )} {\left (3-\sqrt {x^{2}+5}\right )\left (3+\sqrt {x^{2}+5}\right )}\\ &=\lim _{x\to 2}\frac {\left (4-x^{2}\right )\left (3+\sqrt {x^{2}+5}\right )}{9-\left (x^{2}+5\right )}\\ &=\lim _{x\to 2}\frac {\left (4-x^{2}\right )\left (3+\sqrt {x^{2}+5}\right )}{4-x^{2}}\\ &=\lim _{x\to 2}\left (3+\sqrt {x^{2}+5}\right )=3+3=6 . \end {align*}

The factor \(4-x^{2}\) cancelled, which is what made the form determinate.

(d)

\(\lim \limits _{x\longrightarrow 0}\frac {3x}{x^2+2x}\). Substitution gives \(\frac {0}{0}\); factor the denominator and cancel: \[\lim _{x\to 0}\frac {3x}{x(x+2)}=\lim _{x\to 0}\frac {3}{x+2}=\frac {3}{2} .\]