\displaystyle \textbf{Note:}
\displaystyle \lim \limits_{x\to a}\frac{x^m-a^m}{x^n-a^n}
\displaystyle =\lim \limits_{x\to a}\left\{\frac{x^m-a^m}{x-a}\times\frac{x-a}{x^n-a^n}\right\}
\displaystyle =\lim \limits_{x\to a}\left\{\frac{x^m-a^m}{x-a}\div\frac{x^n-a^n}{x-a}\right\}
\displaystyle =\lim \limits_{x\to a}\frac{x^m-a^m}{x-a}\div\lim \limits_{x\to a}\frac{x^n-a^n}{x-a}
\displaystyle =ma^{m-1}\div na^{n-1}
\displaystyle =\frac{m}{n}a^{m-n}
\displaystyle \\

\displaystyle \textbf{Question 1: } \lim \limits_{x\to a}\frac{(x+2)^{\frac{5}{2}}-(a+2)^{\frac{5}{2}}}{x-a}.
\displaystyle \text{Answer:}
\displaystyle \text{Let }y=x+2\text{ and }b=a+2.
\displaystyle \text{When }x\to a,\text{ then }y\to b.
\displaystyle \therefore \lim \limits_{x\to a}\frac{(x+2)^{\frac{5}{2}}-(a+2)^{\frac{5}{2}}}{x-a}=\lim \limits_{y\to b}\frac{y^{\frac{5}{2}}-b^{\frac{5}{2}}}{y-b}.
\displaystyle =\frac{5}{2}b^{\frac{5}{2}-1}
\displaystyle =\frac{5}{2}b^{\frac{3}{2}}
\displaystyle =\frac{5}{2}(a+2)^{\frac{3}{2}}
\displaystyle \\

\displaystyle \textbf{Question 2: } \lim \limits_{x\to a}\frac{(x+2)^{\frac{3}{2}}-(a+2)^{\frac{3}{2}}}{x-a}.
\displaystyle \text{Answer:}
\displaystyle \text{Let }y=x+2\text{ and }b=a+2.
\displaystyle \text{When }x\to a,\text{ then }y\to b.
\displaystyle \therefore \lim \limits_{x\to a}\frac{(x+2)^{\frac{3}{2}}-(a+2)^{\frac{3}{2}}}{x-a}=\lim \limits_{y\to b}\frac{y^{\frac{3}{2}}-b^{\frac{3}{2}}}{y-b}.
\displaystyle =\frac{3}{2}b^{\frac{3}{2}-1}
\displaystyle =\frac{3}{2}b^{\frac{1}{2}}
\displaystyle =\frac{3}{2}(a+2)^{\frac{1}{2}}
\displaystyle \\

\displaystyle \textbf{Question 3: } \lim \limits_{x\to0}\frac{(1+x)^6-1}{(1+x)^2-1}.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle \lim \limits_{x\to0}\frac{(1+x)^6-1}{(1+x)^2-1}=\lim \limits_{x\to0}\left\{\frac{(1+x)^6-1}{x}\times\frac{x}{(1+x)^2-1}\right\}
\displaystyle =\lim \limits_{x\to0}\left\{\frac{(1+x)^6-1^6}{(1+x)-1}\times\frac{(1+x)-1}{(1+x)^2-1^2}\right\}
\displaystyle \text{Let }y=1+x.
\displaystyle \text{When }x\to0,\text{ then }y\to1.
\displaystyle =\lim \limits_{y\to1}\left\{\frac{y^6-1^6}{y-1}\times\frac{y-1}{y^2-1^2}\right\}
\displaystyle =\left(\lim \limits_{y\to1}\frac{y^6-1^6}{y-1}\right)\left(\lim \limits_{y\to1}\frac{y-1}{y^2-1^2}\right)
\displaystyle =6(1)^{5}\times\frac{1}{2(1)}
\displaystyle =3
\displaystyle \\

\displaystyle \textbf{Question 4: } \lim \limits_{x\to a}\frac{x^{\frac{2}{7}}-a^{\frac{2}{7}}}{x-a}.
\displaystyle \text{Answer:}
\displaystyle \lim \limits_{x\to a}\frac{x^{\frac{2}{7}}-a^{\frac{2}{7}}}{x-a}=\frac{2}{7}a^{\frac{2}{7}-1}
\displaystyle =\frac{2}{7}a^{-\frac{5}{7}}
\displaystyle \\

\displaystyle \textbf{Question 5: } \lim \limits_{x\to a}\frac{x^{\frac{5}{7}}-a^{\frac{5}{7}}}{x^{\frac{2}{7}}-a^{\frac{2}{7}}}.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle \lim \limits_{x\to a}\frac{x^{\frac{5}{7}}-a^{\frac{5}{7}}}{x^{\frac{2}{7}}-a^{\frac{2}{7}}}=\left\{\lim \limits_{x\to a}\frac{x^{\frac{5}{7}}-a^{\frac{5}{7}}}{x-a}\times\frac{x-a}{x^{\frac{2}{7}}-a^{\frac{2}{7}}}\right\}
\displaystyle =\left\{\lim \limits_{x\to a}\frac{x^{\frac{5}{7}}-a^{\frac{5}{7}}}{x-a}\div\frac{x^{\frac{2}{7}}-a^{\frac{2}{7}}}{x-a}\right\}
\displaystyle =\frac{5}{7}a^{\frac{5}{7}-1}\div\frac{2}{7}a^{\frac{2}{7}-1}
\displaystyle =\frac{5}{7}a^{-\frac{2}{7}}\div\frac{2}{7}a^{-\frac{5}{7}}
\displaystyle =\frac{5}{2}a^{-\frac{2}{7}-\left(-\frac{5}{7}\right)}
\displaystyle =\frac{5}{2}a^{\frac{3}{7}}
\displaystyle \\

\displaystyle \textbf{Question 6: } \lim \limits_{x\to-\frac{1}{2}}\frac{8x^3+1}{2x+1}.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle \lim \limits_{x\to-\frac{1}{2}}\frac{8x^3+1}{2x+1}=\lim \limits_{x\to-\frac{1}{2}}\frac{(2x)^3-(-1)^3}{2x-(-1)}
\displaystyle \text{Let }y=2x.
\displaystyle \text{When }x\to-\frac{1}{2},\text{ then }y\to-1.
\displaystyle \therefore \lim \limits_{x\to-\frac{1}{2}}\frac{8x^3+1}{2x+1}=\lim \limits_{y\to-1}\frac{y^3-(-1)^3}{y-(-1)}
\displaystyle =3(-1)^{3-1}
\displaystyle =3(-1)^2
\displaystyle =3
\displaystyle \\

\displaystyle \textbf{Question 7: } \lim \limits_{x\to27}\frac{\left(x^{\frac{1}{3}}+3\right)\left(x^{\frac{1}{3}}-3\right)}{x-27}.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle \lim \limits_{x\to27}\frac{\left(x^{\frac{1}{3}}+3\right)\left(x^{\frac{1}{3}}-3\right)}{x-27}
\displaystyle =\lim \limits_{x\to27}\frac{\left(x^{\frac{1}{3}}\right)^2-3^2}{\left(x^{\frac{1}{3}}\right)^3-3^3}
\displaystyle \text{Let }y=x^{\frac{1}{3}}.
\displaystyle \text{When }x\to27,\text{ then }y\to3.
\displaystyle \therefore \lim \limits_{x\to27}\frac{\left(x^{\frac{1}{3}}+3\right)\left(x^{\frac{1}{3}}-3\right)}{x-27}=\lim \limits_{y\to3}\frac{y^2-3^2}{y^3-3^3}
\displaystyle =\frac{2}{3}(3)^{2-3}
\displaystyle =\frac{2}{3}\times\frac{1}{3}
\displaystyle =\frac{2}{9}
\displaystyle \\

\displaystyle \textbf{Question 8: } \lim \limits_{x\to4}\frac{x^3-64}{x^2-16}.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle \lim \limits_{x\to4}\frac{x^3-64}{x^2-16}=\lim \limits_{x\to4}\frac{x^3-4^3}{x^2-4^2}
\displaystyle =\lim \limits_{x\to4}\left\{\frac{x^3-4^3}{x-4}\times\frac{x-4}{x^2-4^2}\right\}
\displaystyle =\lim \limits_{x\to4}\left\{\frac{x^3-4^3}{x-4}\div\frac{x^2-4^2}{x-4}\right\}
\displaystyle =3(4)^{3-1}\div2(4)^{2-1}
\displaystyle =\frac{3\times16}{2\times4}
\displaystyle =6
\displaystyle \\

\displaystyle \textbf{Question 9: } \lim \limits_{x\to1}\frac{x^{15}-1}{x^{10}-1}.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle \lim \limits_{x\to1}\frac{x^{15}-1}{x^{10}-1}=\lim \limits_{x\to1}\left\{\frac{x^{15}-1^{15}}{x-1}\times\frac{x-1}{x^{10}-1^{10}}\right\}
\displaystyle =\lim \limits_{x\to1}\left\{\frac{x^{15}-1^{15}}{x-1}\div\frac{x^{10}-1^{10}}{x-1}\right\}
\displaystyle =15(1)^{15-1}\div10(1)^{10-1}
\displaystyle =\frac{15}{10}
\displaystyle =\frac{3}{2}
\displaystyle \\

\displaystyle \textbf{Question 10: } \lim \limits_{x\to-1}\frac{x^3+1}{x+1}.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle \lim \limits_{x\to-1}\frac{x^3+1}{x+1}=\lim \limits_{x\to-1}\frac{x^3-(-1)^3}{x-(-1)}
\displaystyle =3(-1)^{3-1}
\displaystyle =3
\displaystyle \\

\displaystyle \textbf{Question 11: } \lim \limits_{x\to a}\frac{x^{\frac{2}{3}}-a^{\frac{2}{3}}}{x^{\frac{3}{4}}-a^{\frac{3}{4}}}.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle \lim \limits_{x\to a}\frac{x^{\frac{2}{3}}-a^{\frac{2}{3}}}{x^{\frac{3}{4}}-a^{\frac{3}{4}}}=\lim \limits_{x\to a}\frac{x^{\frac{2}{3}}-a^{\frac{2}{3}}}{x-a}\times\frac{x-a}{x^{\frac{3}{4}}-a^{\frac{3}{4}}}
\displaystyle =\lim \limits_{x\to a}\frac{x^{\frac{2}{3}}-a^{\frac{2}{3}}}{x-a}\div\frac{x^{\frac{3}{4}}-a^{\frac{3}{4}}}{x-a}
\displaystyle =\frac{\frac{2}{3}a^{\frac{2}{3}-1}}{\frac{3}{4}a^{\frac{3}{4}-1}}
\displaystyle =\frac{8}{9}a^{-\frac{1}{3}-\left(-\frac{1}{4}\right)}
\displaystyle =\frac{8}{9}a^{-\frac{1}{12}}
\displaystyle \\

\displaystyle \textbf{Question 12: } \text{If }\lim \limits_{x\to3}\frac{x^n-3^n}{x-3}=108,\text{ find the value of }n.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\lim \limits_{x\to3}\frac{x^n-3^n}{x-3}=108
\displaystyle \therefore n(3)^{n-1}=108
\displaystyle \Rightarrow n(3)^{n-1}=4(3)^3
\displaystyle \text{On putting }n=4,\text{ we get }4(3)^3=4\times27=108.
\displaystyle \therefore n=4
\displaystyle \\

\displaystyle \textbf{Question 13: } \text{If }\lim \limits_{x\to a}\frac{x^9-a^9}{x-a}=9,\text{ find the value of }a.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\lim \limits_{x\to a}\frac{x^9-a^9}{x-a}=9
\displaystyle \therefore 9a^{9-1}=9
\displaystyle \Rightarrow 9a^8=9
\displaystyle \Rightarrow a^8=1
\displaystyle \Rightarrow a=\pm1
\displaystyle \\

\displaystyle \textbf{Question 14: } \text{If }\lim \limits_{x\to a}\frac{x^5-a^5}{x-a}=405,\text{ find the value of }a.
\displaystyle \text{Answer:}
\displaystyle \text{Given }\lim \limits_{x\to a}\frac{x^5-a^5}{x-a}=405
\displaystyle \Rightarrow 5a^{5-1}=405
\displaystyle \Rightarrow 5a^4=405
\displaystyle \Rightarrow a^4=81
\displaystyle \Rightarrow a=\pm3
\displaystyle \\

\displaystyle \textbf{Question 15: } \text{If }\lim \limits_{x\to a}\frac{x^9-a^9}{x-a}=\lim \limits_{x\to5}(4+x),\text{ find the value of }a.
\displaystyle \text{Answer:}
\displaystyle \text{Given }\lim \limits_{x\to a}\frac{x^9-a^9}{x-a}=\lim \limits_{x\to5}(4+x)
\displaystyle \Rightarrow 9a^{9-1}=9
\displaystyle \Rightarrow 9a^8=9
\displaystyle \Rightarrow a^8=1
\displaystyle \Rightarrow a=\pm1
\displaystyle \\

\displaystyle \textbf{Question 16: } \text{If }\lim \limits_{x\to a}\frac{x^3-a^3}{x-a}=\lim \limits_{x\to1}\frac{x^4-1}{x-1},\text{ find the value of }a.
\displaystyle \text{Answer:}
\displaystyle \text{Given }\lim \limits_{x\to a}\frac{x^3-a^3}{x-a}=\lim \limits_{x\to1}\frac{x^4-1}{x-1}
\displaystyle \Rightarrow 3a^{3-1}=4(1)^{4-1}
\displaystyle \Rightarrow 3a^2=4
\displaystyle \Rightarrow a^2=\frac{4}{3}
\displaystyle \Rightarrow a=\pm\frac{2}{\sqrt3}
\displaystyle \\

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