Given,

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

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

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

Given,

$$\begin{aligned} & \lim _{x \rightarrow-3} \frac{x^3+27}{x^5+243}=\lim _{x \rightarrow-3} \frac{\frac{x^3+27}{x+3}}{\frac{x^5+243}{x+3}} \\ & =\lim _{x \rightarrow-3} \frac{\frac{x^3-(-3)^3}{x-(-3)}}{\frac{x^5-(-3)^5}{x-(-3)}}=\frac{\lim _\limits{x \rightarrow-3} \frac{x^3-(-3)^3}{x-(-3)}}{\lim _\limits{x \rightarrow-3} \frac{x^5-(-3)^5}{x-(-3)}} \quad\left[\because \lim _{x \rightarrow a} \frac{f(x)}{g(x)}=\frac{\lim _{x \rightarrow \mathrm{a}} f(x)}{\lim _{x \rightarrow a} g(x)}\right] \\ & =\frac{3(-3)^{3-1}}{5(-3)^{5-1}}=\frac{3}{5} \frac{(-3)^2}{(-3)^4} \\ & {\left[\because \lim _{x \rightarrow a} \frac{x^n-a^n}{x-a}=n a^{n-1}\right]} \\ & =\frac{3}{5(-3)^2}=\frac{3}{45}=\frac{1}{15} \end{aligned}$$

Given, $$\begin{aligned} \lim _{x \rightarrow 1 / 2}\left(\frac{8 x-3}{2 x-1}-\frac{4 x^2+1}{4 x^2-1}\right) & =\lim _{x \rightarrow 1 / 2}\left[\frac{(8 x-3)(2 x+1)-\left(4 x^2+1\right)}{\left(4 x^2-1\right)}\right] \\ & =\lim _{x \rightarrow 1 / 2}\left[\frac{16 x^2+8 x-6 x-3-4 x^2-1}{4 x^2-1}\right] \\ & =\lim _{x \rightarrow 1 / 2}\left[\frac{12 x^2+2 x-4}{4 x^2-1}\right] \\ & =\lim _{x \rightarrow 1 / 2} \frac{2\left(6 x^2+x-2\right)}{4 x^2-1} \end{aligned}$$

$$\begin{aligned} & =\lim _{x \rightarrow 1 / 2} \frac{2\left(6 x^2+4 x-3 x-2\right)}{4 x^2-1} \\ & =\lim _{x \rightarrow 1 / 2} \frac{2[2 x(3 x+2)-1(3 x+2)]}{4 x^2-1} \\ & =\lim _{x \rightarrow 1 / 2} \frac{2[(3 x+2)(2 x-1)]}{(2 x)^2-(1)^2} \\ & =\lim _{x \rightarrow 1 / 2} \frac{2(3 x+2)(2 x-1)}{(2 x-1)(2 x+1)} \\ & =\lim _{x \rightarrow 1 / 2} \frac{2(3 x+2)}{2 x-1}=\frac{2\left(3 \times \frac{1}{2}+2\right)}{2 \times \frac{1}{2}+1} \\ & =\frac{3}{2}+2=\frac{7}{2} \end{aligned}$$

$$\begin{aligned} \text { Given, } & \lim _{x \rightarrow 2} \frac{x^n-2^n}{x-2} =80 \\ \Rightarrow \quad & n(2)^{n-1} =80 \quad \left[\because \lim _{x \rightarrow a} \frac{x^n-a^n}{x-a}=n a^{n-1}\right]\\ \Rightarrow \quad & n(2)^{n-1} =5 \times 16 \\ \Rightarrow \quad & n \times 2^{n-1} =5 \times(2)^4 \\ \Rightarrow \quad & n \times 2^{n-1} =5 \times(2)^{5-1} \\ \therefore \quad & n =5 \end{aligned} $$