Given,

$$\begin{aligned} \lim _{x \rightarrow 0} \frac{(x+2)^{1 / 3}-2^{1 / 3}}{x} & =\lim _{x \rightarrow 0} \frac{(x+2)^{1 / 3}-2^{1 / 3}}{(x+2)-2} \\ & =\frac{1}{3} \times 2^{\frac{1}{3}-1} \\ & =\frac{1}{3} \times(2)^{-2 / 3} \quad \left[\because \lim _{x \rightarrow a} \frac{x^n-a^n}{x-a}=n a^{n-1}\right]\\ & =\frac{1}{3(2)^{2 / 3}} \quad [\because x \rightarrow 0 \Rightarrow x+2 \rightarrow 2] \end{aligned}$$

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 $x$ ]

$$\begin{array}{ll} =\lim _\limits{x \rightarrow 0} \frac{\frac{(1+x)^6-1}{(1+x)-1}}{\frac{(1+x)^2-1}{(1+x)-1}} & {[\because x \rightarrow 0 \Rightarrow 1+x \rightarrow 1]} \\ =\frac{\lim _\limits{x \rightarrow 0} \frac{(1+x)^6-(1)^6}{(1+x)-1}}{\lim _\limits{x \rightarrow 0} \frac{(1+x)^2-(1)^2}{(1+x)-1}} & {\left[\because \lim _\limits{x \rightarrow a} \frac{f(x)}{g(x)}=\frac{\lim _\limits{x \rightarrow a} f(x)}{\lim _\limits{x \rightarrow a} g(x)}\right]} \\ =\frac{6(1)^{6-1}}{2(1)^{2-1}} & {\left[\therefore \lim _\limits{x \rightarrow a} \frac{x^n-a^n}{x-a}=n a^{n-1}\right]} \\ =\frac{6 \times 1}{2 \times 1}=\frac{6}{2}=3 & \end{array}$$

Given,

$$\begin{aligned} \lim _{x \rightarrow a} \frac{(2+x)^{5 / 2}-(a+2)^{5 / 2}}{x-a} & =\lim _{x \rightarrow a} \frac{(2+x)^{5 / 2}-(a+2)^{5 / 2}}{(2+x)-(a+2)} & \\ & =\frac{5}{2}(a+2)^{\frac{5}{2}-1} & {\left[\because \lim _{x \rightarrow a} \frac{x^n-a^n}{x-a}=n a^{n-1}\right] } \\ & =\frac{5}{2}(a+2)^{3 / 2} & {[\because x \rightarrow a \Rightarrow x+2 \rightarrow a+2] } \end{aligned}$$

Given, $$ \begin{aligned} \lim _{x \rightarrow 1} \frac{x^4-\sqrt{x}}{\sqrt{x}-1}= & \lim _{x \rightarrow 1} \frac{\sqrt{x}\left[(x)^{7 / 2}-1\right]}{\sqrt{x}-1} \\ & =\lim _{x \rightarrow 1} \frac{(x)^{7 / 2}-1}{\sqrt{x}-1} \cdot \lim _{x \rightarrow 1} \sqrt{x} \quad\left[\because \lim _{x \rightarrow a} f(x) \cdot g(x)=\lim _{x \rightarrow a} f(x) \cdot \lim _{x \rightarrow a} g(x)\right] \\ & =\lim _{x \rightarrow 1} \frac{\frac{x^{7 / 2}-1}{x-1}}{\frac{(x)^{1 / 2}-1}{x-1}} \cdot 1 \\ & =\frac{\lim _\limits{x \rightarrow 1} \frac{x^{7 / 2}-1}{x-1}}{\lim _\limits{x \rightarrow 1} \frac{(x)^{1 / 2}-1}{x-1}} \quad\left[\because \lim _{x \rightarrow a} \frac{f(x)}{g(x)}=\frac{\lim _{x \rightarrow a} f(x)}{\lim _\limits{x \rightarrow a} g(x)}\right] \\ & =\frac{\frac{7}{2}(1)^{\frac{7}{2}-1}}{\frac{1}{2}(1)^{\frac{1}{2}-1}}=\frac{\frac{7}{2}}{\frac{1}{2}}=7 \end{aligned}$$

Given,

$\begin{aligned} \lim _{x \rightarrow 2} \frac{x^2-4}{\sqrt{3 x-2}-\sqrt{x+2}} & =\lim _{x \rightarrow 2} \frac{\left.\left(x^2-4\right) \sqrt{3 x-2}+\sqrt{x+2}\right)}{(\sqrt{3 x-2}-\sqrt{x+2})(\sqrt{3 x-2}-\sqrt{x+2})} \\ & =\lim _{x \rightarrow 2} \frac{\left(x^2-4\right)(\sqrt{3 x-2}+\sqrt{x+2})}{(\sqrt{3 x-2})^2-(\sqrt{x+2})^2} \quad [\because (a+b)(a-b)=a^2-b^2]\\ & =\lim _{x \rightarrow 2} \frac{\left(x^2-4\right)(\sqrt{3 x-2}+\sqrt{x+2})}{(3 x-2)-(x+2)} \\ & =\lim _{x \rightarrow 2} \frac{\left(x^2-4\right)(\sqrt{3 x-2}+\sqrt{x+2})}{3 x-2-x-2} \\ & =\lim _{x \rightarrow 2} \frac{\left(x^2-4\right)(\sqrt{3 x-2}+\sqrt{x+2})}{2 x-4} \\ & =\lim _{x \rightarrow 2} \frac{(x+2)(x-2)(\sqrt{3 x-2}+\sqrt{x+2})}{2(x-2)} \\ & =\lim _{x \rightarrow 2} \frac{(x+2)(\sqrt{3 x-2}+\sqrt{x+2})}{2} \\ & =\frac{(2+2)(\sqrt{6-2}+\sqrt{2+2})}{2} \\ & =\frac{4(2+2)}{2}=8\end{aligned}$