Evaluate $\lim _{x \rightarrow 3} \frac{x^2-9}{x-3}$.
Given,
$$\begin{aligned} \lim _{x \rightarrow 3} \frac{x^2-9}{x-3} & =\lim _{x \rightarrow 3} \frac{x^2-(3)^2}{x-3} \\ & =\lim _{x \rightarrow 3} \frac{(x+3)(x-3)}{(x-3)}=\lim _{x \rightarrow 3}(x+3) \\ & =3+3=6 \end{aligned}$$
Evaluate $\lim _\limits{x \rightarrow 1 / 2} \frac{4 x^2-1}{2 x-1}$.
Given,
$$\begin{aligned} \lim _{x \rightarrow 1 / 2} \frac{4 x^2-1}{2 x-1} & =\lim _{x \rightarrow 1 / 2} \frac{(2 x)^2-(1)^2}{2 x-1} \\ & =\lim _{x \rightarrow 1 / 2} \frac{(2 x+1)(2 x-1)}{(2 x-1)}=\lim _{x \rightarrow 1 / 2}(2 x+1) \\ & =2 \times \frac{1}{2}+1=1+1=2 \end{aligned}$$
Evaluate $\lim _\limits{h \rightarrow 0} \frac{\sqrt{x+h}-\sqrt{x}}{h}$.
Given,
$$\begin{aligned} \lim _{h \rightarrow 0} \frac{\sqrt{x+h}-\sqrt{x}}{h} & =\lim _{h \rightarrow 0} \frac{(x+h)^{1 / 2}-(x)^{1 / 2}}{x+h-x} & \\ & =\lim _{h \rightarrow 0} \frac{(x+h)^{1 / 2}-(x)^{1 / 2}}{(x+h)-x} & {\left[\because \lim _{x \rightarrow a} \frac{x^n-a^n}{x-a}=n a^{n-1}\right] } \\ & =\frac{1}{2} x^{\frac{1}{2}-1}=\frac{1}{2} x^{-1 / 2} & {[\because h \rightarrow 0 \Rightarrow x+h \rightarrow x] } \\ & =\frac{1}{2 \sqrt{x}} & \end{aligned}$$
Evaluate $\lim _\limits{x \rightarrow 0} \frac{(x+2)^{1 / 3}-2^{1 / 3}}{x}$.
Given,
$$\begin{aligned} \lim _{x \rightarrow 0} \frac{(x+2)^{1 / 3}-2^{1 / 3}}{x} & =\lim _{x \rightarrow 0} \frac{(x+2)^{1 / 3}-2^{1 / 3}}{(x+2)-2} \\ & =\frac{1}{3} \times 2^{\frac{1}{3}-1} \\ & =\frac{1}{3} \times(2)^{-2 / 3} \quad \left[\because \lim _{x \rightarrow a} \frac{x^n-a^n}{x-a}=n a^{n-1}\right]\\ & =\frac{1}{3(2)^{2 / 3}} \quad [\because x \rightarrow 0 \Rightarrow x+2 \rightarrow 2] \end{aligned}$$
Evaluate $\lim _\limits{x \rightarrow 0} \frac{(1+x)^6-1}{(1+x)^2-1}$.
Given, $\lim _\limits{x \rightarrow 0} \frac{(1+x)^6-1}{(1+x)^2-1}=\lim _\limits{x \rightarrow 0} \frac{\frac{(1+x)^6-1}{x}}{\frac{(1+x)^2-1}{x}}\quad$ [dividing numerator and denominator by $x$ ]
$$\begin{array}{ll} =\lim _\limits{x \rightarrow 0} \frac{\frac{(1+x)^6-1}{(1+x)-1}}{\frac{(1+x)^2-1}{(1+x)-1}} & {[\because x \rightarrow 0 \Rightarrow 1+x \rightarrow 1]} \\ =\frac{\lim _\limits{x \rightarrow 0} \frac{(1+x)^6-(1)^6}{(1+x)-1}}{\lim _\limits{x \rightarrow 0} \frac{(1+x)^2-(1)^2}{(1+x)-1}} & {\left[\because \lim _\limits{x \rightarrow a} \frac{f(x)}{g(x)}=\frac{\lim _\limits{x \rightarrow a} f(x)}{\lim _\limits{x \rightarrow a} g(x)}\right]} \\ =\frac{6(1)^{6-1}}{2(1)^{2-1}} & {\left[\therefore \lim _\limits{x \rightarrow a} \frac{x^n-a^n}{x-a}=n a^{n-1}\right]} \\ =\frac{6 \times 1}{2 \times 1}=\frac{6}{2}=3 & \end{array}$$