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,

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

Given, $$ \begin{aligned} \lim _{x \rightarrow \sqrt{2}} \frac{x^4-4}{x^2+3 \sqrt{2} x-8} & =\lim _{x \rightarrow \sqrt{2}} \frac{\left(x^2\right)^2-(2)^2}{x^2+3 \sqrt{2} x-8} \\ & =\lim _{x \rightarrow \sqrt{2}} \frac{\left(x^2-2\right)\left(x^2+2\right)}{x^2+4 \sqrt{2} x-\sqrt{2} x-8} \\ & =\lim _{x \rightarrow \sqrt{2}} \frac{(x-\sqrt{2})(x+\sqrt{2})\left(x^2+2\right)}{x(x+4 \sqrt{2})-\sqrt{2}(x+4 \sqrt{2})} \\ & =\lim _{x \rightarrow \sqrt{2}} \frac{(x-\sqrt{2})(x+\sqrt{2})\left(x^2+2\right)}{(x-\sqrt{2})(x+4 \sqrt{2})} \\ & =\lim _{x \rightarrow \sqrt{2}} \frac{(x+\sqrt{2})\left(x^2+2\right)}{(x+4 \sqrt{2})} \\ & =\frac{(\sqrt{2}+\sqrt{2})\left[(\sqrt{2})^2+2\right]}{(\sqrt{2}+4 \sqrt{2})} \\ & =\frac{2 \sqrt{2}(2+2)}{5 \sqrt{2}}=\frac{8}{5} \end{aligned} $$

Given,

$$\begin{aligned} & \lim _{x \rightarrow 1} \frac{x^7-2 x^5+1}{x^3-3 x^2+2} \quad \left[\frac{0}{0}\text{form}\right]\\ = & \lim _{x \rightarrow 1} \frac{x^7-x^5-x^5+1}{x^3-x^2-2 x^2+2} \\ = & \lim _{x \rightarrow 1} \frac{x^5\left(x^2-1\right)-1\left(x^5-1\right)}{x^2(x-1)-2\left(x^2-1\right)} \end{aligned}$$

$$\begin{aligned} &\text { On dividing numerator and denominator by }(x-1) \text {, then }\\ &\begin{aligned} & =\lim _{x \rightarrow 1} \frac{\frac{x^5\left(x^2-1\right)}{(x-1)}-\frac{1\left(x^2-1\right)}{(x-1)}}{\frac{x^2(x-1)}{(x-1)}-\frac{2\left(x^2-1\right)}{(x-1)}} \\ & =\frac{\lim _\limits{x \rightarrow 1} x^5(x+1)-\lim _\limits{x \rightarrow 1}\left(\frac{x^5-1}{x-1}\right)}{\lim _\limits{x \rightarrow 1} x^2-\lim _\limits{x \rightarrow 1}(x+1)} \\ & =\frac{1 \times 2-5 \times(1)^4}{1-2 \times 2}=\frac{2-5}{1-4} \\ & =\frac{-3}{-3}=1 \end{aligned} \end{aligned}$$

$$\begin{aligned} &\text { Given, }\\ &\begin{aligned} \lim _{x \rightarrow 0} \frac{\sqrt{1+x^3}-\sqrt{1-x^3}}{x^2} & =\lim _{x \rightarrow 0} \frac{\sqrt{1+x^3}-\sqrt{1-x^3}}{x^2} \cdot \frac{\sqrt{1+x^3}+\sqrt{1-x^3}}{\sqrt{1+x^3}+\sqrt{1-x^3}} \\ & =\lim _{x \rightarrow 0} \frac{\left(1+x^3\right)-\left(1-x^3\right)}{x^2\left(\sqrt{1+x^3}+\sqrt{1-x^3}\right)} \\ & =\lim _{x \rightarrow 0} \frac{1+x^3-1+x^3}{x^2\left(\sqrt{1+x^3}+\sqrt{1-x^3}\right)} \\ & =\lim _{x \rightarrow 0} \frac{2 x^3}{x^2\left(\sqrt{1+x^3}+\sqrt{1-x^3}\right)} \\ & =\lim _{x \rightarrow 0} \frac{2 x}{\left(\sqrt{1+x^3}+\sqrt{1-x^3}\right)} \\ & =0 \end{aligned} \end{aligned}$$