\displaystyle \textbf{Question 1: }\text{Show that }\lim\limits_{x\to0}\frac{x}{|x|}\text{ does not exist.}
\displaystyle \text{Answer:}
\displaystyle \text{Left-hand limit:}
\displaystyle \lim\limits_{x\to0^-}\frac{x}{|x|}
\displaystyle \text{Put }x=0-h,\text{ where }h\to0^+.
\displaystyle \lim\limits_{x\to0^-}\frac{x}{|x|}=\lim\limits_{h\to0^+}\frac{-h}{|-h|}
\displaystyle =\lim\limits_{h\to0^+}\frac{-h}{h}=-1.
\displaystyle \text{Right-hand limit:}
\displaystyle \lim\limits_{x\to0^+}\frac{x}{|x|}
\displaystyle \text{Put }x=0+h,\text{ where }h\to0^+.
\displaystyle \lim\limits_{x\to0^+}\frac{x}{|x|}=\lim\limits_{h\to0^+}\frac{h}{|h|}
\displaystyle =\lim\limits_{h\to0^+}\frac{h}{h}=1.
\displaystyle \therefore \lim\limits_{x\to0^-}\frac{x}{|x|}\neq\lim\limits_{x\to0^+}\frac{x}{|x|}.
\displaystyle \therefore \lim\limits_{x\to0}\frac{x}{|x|}\text{ does not exist.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Find }k\text{ so that }\lim\limits_{x\to2}f(x)\text{ may exist, where}
\displaystyle f(x)=\left\{\begin{array}{ll}2x+3,&x\le2\\x+k,&x>2\end{array}\right.
\displaystyle \text{Answer:}
\displaystyle \text{Left-hand limit at }x=2:
\displaystyle \lim\limits_{x\to2^-}f(x)=\lim\limits_{x\to2^-}(2x+3).
\displaystyle \text{Put }x=2-h,\text{ where }h\to0^+.
\displaystyle \lim\limits_{x\to2^-}(2x+3)=\lim\limits_{h\to0^+}[2(2-h)+3]=7.
\displaystyle \text{Right-hand limit at }x=2:
\displaystyle \lim\limits_{x\to2^+}f(x)=\lim\limits_{x\to2^+}(x+k).
\displaystyle \text{Put }x=2+h,\text{ where }h\to0^+.
\displaystyle \lim\limits_{x\to2^+}(x+k)=\lim\limits_{h\to0^+}(2+h+k)=2+k.
\displaystyle \text{For }\lim\limits_{x\to2}f(x)\text{ to exist, LHL = RHL.}
\displaystyle \therefore 7=2+k.
\displaystyle \therefore k=5.
\displaystyle \text{Hence, }\lim\limits_{x\to2}f(x)\text{ exists when }k=5.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Show that }\lim\limits_{x\to0}\frac{1}{x}\text{ does not exist.}
\displaystyle \text{Answer:}
\displaystyle \text{Left-hand limit at }x=0:
\displaystyle \lim\limits_{x\to0^-}\frac{1}{x}
\displaystyle \text{Put }x=0-h,\text{ where }h\to0^+.
\displaystyle \lim\limits_{x\to0^-}\frac{1}{x}=\lim\limits_{h\to0^+}\frac{1}{-h}=-\infty.
\displaystyle \text{Right-hand limit at }x=0:
\displaystyle \lim\limits_{x\to0^+}\frac{1}{x}
\displaystyle \text{Put }x=0+h,\text{ where }h\to0^+.
\displaystyle \lim\limits_{x\to0^+}\frac{1}{x}=\lim\limits_{h\to0^+}\frac{1}{h}=+\infty.
\displaystyle \therefore \lim\limits_{x\to0^-}\frac{1}{x}\neq\lim\limits_{x\to0^+}\frac{1}{x}.
\displaystyle \therefore \lim\limits_{x\to0}\frac{1}{x}\text{ does not exist.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Let }f(x)\text{ be defined by}
\displaystyle f(x)=\left\{\begin{array}{ll}\dfrac{3x}{|x|+2x},&x\neq0\\0,&x=0\end{array}\right.
\displaystyle \text{Show that }\lim\limits_{x\to0}f(x)\text{ does not exist.}
\displaystyle \text{Answer:}
\displaystyle \text{Left-hand limit at }x=0:
\displaystyle \lim\limits_{x\to0^-}f(x)=\lim\limits_{x\to0^-}\frac{3x}{|x|+2x}.
\displaystyle \text{Put }x=0-h,\text{ where }h\to0^+.
\displaystyle \lim\limits_{x\to0^-}f(x)=\lim\limits_{h\to0^+}\frac{3(-h)}{|-h|+2(-h)}
\displaystyle =\lim\limits_{h\to0^+}\frac{-3h}{h-2h}=\lim\limits_{h\to0^+}\frac{-3h}{-h}=3.
\displaystyle \text{Right-hand limit at }x=0:
\displaystyle \lim\limits_{x\to0^+}f(x)=\lim\limits_{x\to0^+}\frac{3x}{|x|+2x}.
\displaystyle \text{Put }x=0+h,\text{ where }h\to0^+.
\displaystyle \lim\limits_{x\to0^+}f(x)=\lim\limits_{h\to0^+}\frac{3h}{|h|+2h}
\displaystyle =\lim\limits_{h\to0^+}\frac{3h}{h+2h}=\lim\limits_{h\to0^+}\frac{3h}{3h}=1.
\displaystyle \therefore \lim\limits_{x\to0^-}f(x)\neq\lim\limits_{x\to0^+}f(x).
\displaystyle \therefore \lim\limits_{x\to0}f(x)\text{ does not exist.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Let }f(x)=\left\{\begin{array}{ll}x+1,&x>0\\x-1,&x<0\end{array}\right.
\displaystyle \text{Prove that }\lim\limits_{x\to0}f(x)\text{ does not exist.}
\displaystyle \text{Answer:}
\displaystyle \text{Left-hand limit at }x=0:
\displaystyle \lim\limits_{x\to0^-}f(x)=\lim\limits_{x\to0^-}(x-1).
\displaystyle \text{Put }x=0-h,\text{ where }h\to0^+.
\displaystyle \lim\limits_{x\to0^-}(x-1)=\lim\limits_{h\to0^+}(0-h-1)=-1.
\displaystyle \text{Right-hand limit at }x=0:
\displaystyle \lim\limits_{x\to0^+}f(x)=\lim\limits_{x\to0^+}(x+1).
\displaystyle \text{Put }x=0+h,\text{ where }h\to0^+.
\displaystyle \lim\limits_{x\to0^+}(x+1)=\lim\limits_{h\to0^+}(0+h+1)=1.
\displaystyle \therefore \lim\limits_{x\to0^-}f(x)\neq\lim\limits_{x\to0^+}f(x).
\displaystyle \therefore \lim\limits_{x\to0}f(x)\text{ does not exist.}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Let }f(x)=\left\{\begin{array}{ll}x+5,&x>0\\x-4,&x<0\end{array}\right.
\displaystyle \text{Prove that }\lim\limits_{x\to0}f(x)\text{ does not exist.}
\displaystyle \text{Answer:}
\displaystyle \text{Left-hand limit at }x=0:
\displaystyle \lim\limits_{x\to0^-}f(x)=\lim\limits_{x\to0^-}(x-4).
\displaystyle \text{Put }x=0-h,\text{ where }h\to0^+.
\displaystyle \lim\limits_{x\to0^-}(x-4)=\lim\limits_{h\to0^+}(0-h-4)=-4.
\displaystyle \text{Right-hand limit at }x=0:
\displaystyle \lim\limits_{x\to0^+}f(x)=\lim\limits_{x\to0^+}(x+5).
\displaystyle \text{Put }x=0+h,\text{ where }h\to0^+.
\displaystyle \lim\limits_{x\to0^+}(x+5)=\lim\limits_{h\to0^+}(0+h+5)=5.
\displaystyle \therefore \lim\limits_{x\to0^-}f(x)\neq\lim\limits_{x\to0^+}f(x).
\displaystyle \therefore \lim\limits_{x\to0}f(x)\text{ does not exist.}
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Find }\lim\limits_{x\to3}f(x)\text{, where }f(x)=\left\{\begin{array}{ll}4,&x>3\\x+1,&x<3\end{array}\right.
\displaystyle \text{Answer:}
\displaystyle \text{Left-hand limit at }x=3:
\displaystyle \lim\limits_{x\to3^-}f(x)=\lim\limits_{x\to3^-}(x+1).
\displaystyle \text{Put }x=3-h,\text{ where }h\to0^+.
\displaystyle \lim\limits_{x\to3^-}(x+1)=\lim\limits_{h\to0^+}(3-h+1)=4.
\displaystyle \text{Right-hand limit at }x=3:
\displaystyle \lim\limits_{x\to3^+}f(x)=\lim\limits_{x\to3^+}4.
\displaystyle \text{Put }x=3+h,\text{ where }h\to0^+.
\displaystyle \lim\limits_{x\to3^+}4=4.
\displaystyle \therefore \lim\limits_{x\to3^-}f(x)=\lim\limits_{x\to3^+}f(x)=4.
\displaystyle \therefore \lim\limits_{x\to3}f(x)=4.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{If }f(x)=\left\{\begin{array}{ll}2x+3,&x\le0\\3(x+1),&x>0\end{array}\right.
\displaystyle \text{Find }\lim\limits_{x\to0}f(x)\text{ and }\lim\limits_{x\to1}f(x).
\displaystyle \text{Answer:}
\displaystyle \text{Left-hand limit at }x=0:
\displaystyle \lim\limits_{x\to0^-}f(x)=\lim\limits_{x\to0^-}(2x+3).
\displaystyle \text{Put }x=0-h,\text{ where }h\to0^+.
\displaystyle \lim\limits_{x\to0^-}(2x+3)=\lim\limits_{h\to0^+}[2(0-h)+3]=\lim\limits_{h\to0^+}(-2h+3)=3.
\displaystyle \text{Right-hand limit at }x=0:
\displaystyle \lim\limits_{x\to0^+}f(x)=\lim\limits_{x\to0^+}3(x+1).
\displaystyle \text{Put }x=0+h,\text{ where }h\to0^+.
\displaystyle \lim\limits_{x\to0^+}3(x+1)=\lim\limits_{h\to0^+}3[(0+h)+1]=\lim\limits_{h\to0^+}(3h+3)=3.
\displaystyle \therefore \lim\limits_{x\to0^-}f(x)=\lim\limits_{x\to0^+}f(x)=3.
\displaystyle \therefore \lim\limits_{x\to0}f(x)=3.

\displaystyle \text{Left-hand limit at }x=1:
\displaystyle \lim\limits_{x\to1^-}f(x)=\lim\limits_{x\to1^-}3(x+1).
\displaystyle \text{Put }x=1-h,\text{ where }h\to0^+.
\displaystyle \lim\limits_{x\to1^-}3(x+1)=\lim\limits_{h\to0^+}3(1-h+1)=\lim\limits_{h\to0^+}(6-3h)=6.
\displaystyle \text{Right-hand limit at }x=1:
\displaystyle \lim\limits_{x\to1^+}f(x)=\lim\limits_{x\to1^+}3(x+1).
\displaystyle \text{Put }x=1+h,\text{ where }h\to0^+.
\displaystyle \lim\limits_{x\to1^+}3(x+1)=\lim\limits_{h\to0^+}3(1+h+1)=\lim\limits_{h\to0^+}(6+3h)=6.
\displaystyle \therefore \lim\limits_{x\to1^-}f(x)=\lim\limits_{x\to1^+}f(x)=6.
\displaystyle \therefore \lim\limits_{x\to1}f(x)=6.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{Find }\lim\limits_{x\to1}f(x),\text{ if }f(x)=\left\{\begin{array}{ll}x^2-1,&x\le1\\-x^2-1,&x>1\end{array}\right.
\displaystyle \text{Answer:}
\displaystyle \text{Left-hand limit at }x=1:
\displaystyle \lim\limits_{x\to1^-}f(x)=\lim\limits_{x\to1^-}(x^2-1).
\displaystyle \text{Put }x=1-h,\text{ where }h\to0^+.
\displaystyle \lim\limits_{x\to1^-}(x^2-1)=\lim\limits_{h\to0^+}\left[(1-h)^2-1\right].
\displaystyle \lim\limits_{h\to0^+}(1-2h+h^2-1)=\lim\limits_{h\to0^+}(-2h+h^2)=0.
\displaystyle \text{Right-hand limit at }x=1:
\displaystyle \lim\limits_{x\to1^+}f(x)=\lim\limits_{x\to1^+}(-x^2-1).
\displaystyle \text{Put }x=1+h,\text{ where }h\to0^+.
\displaystyle \lim\limits_{x\to1^+}(-x^2-1)=\lim\limits_{h\to0^+}\left[-(1+h)^2-1\right].
\displaystyle \lim\limits_{h\to0^+}(-1-2h-h^2-1)=\lim\limits_{h\to0^+}(-2-2h-h^2)=-2.
\displaystyle \therefore \lim\limits_{x\to1^-}f(x)\neq\lim\limits_{x\to1^+}f(x).
\displaystyle \therefore \lim\limits_{x\to1}f(x)\text{ does not exist.}
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{Evaluate }\lim\limits_{x\to0}f(x)\text{ where }f(x)=\left\{\begin{array}{ll}\dfrac{|x|}{x},&x\neq0\\0,&x=0\end{array}\right.
\displaystyle \text{Answer:}
\displaystyle \text{Left-hand limit at }x=0:
\displaystyle \lim\limits_{x\to0^-}f(x)=\lim\limits_{x\to0^-}\frac{|x|}{x}.
\displaystyle \text{Put }x=0-h,\text{ where }h\to0^+.
\displaystyle \lim\limits_{x\to0^-}\frac{|x|}{x}=\lim\limits_{h\to0^+}\frac{|0-h|}{0-h}=\lim\limits_{h\to0^+}\frac{h}{-h}=-1.
\displaystyle \text{Right-hand limit at }x=0:
\displaystyle \lim\limits_{x\to0^+}f(x)=\lim\limits_{x\to0^+}\frac{|x|}{x}.
\displaystyle \text{Put }x=0+h,\text{ where }h\to0^+.
\displaystyle \lim\limits_{x\to0^+}\frac{|x|}{x}=\lim\limits_{h\to0^+}\frac{|0+h|}{0+h}=\lim\limits_{h\to0^+}\frac{h}{h}=1.
\displaystyle \therefore \lim\limits_{x\to0^-}f(x)\neq\lim\limits_{x\to0^+}f(x).
\displaystyle \therefore \lim\limits_{x\to0}f(x)\text{ does not exist.}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{Let }a_1,a_2,\ldots,a_n\text{ be fixed real numbers such that}
\displaystyle f(x)=(x-a_1)(x-a_2)\cdots(x-a_n).
\displaystyle \text{What is }\lim\limits_{x\to a_1}f(x)?\text{ For }a\neq a_1,a_2,\ldots,a_n,\text{ compute}
\displaystyle \lim\limits_{x\to a}f(x).
\displaystyle \text{Answer:}
\displaystyle \text{Since }f(x)\text{ is a polynomial, its limit can be found by direct substitution.}
\displaystyle \lim\limits_{x\to a_1}f(x)=(a_1-a_1)(a_1-a_2)\cdots(a_1-a_n)
\displaystyle =0.
\displaystyle \therefore \lim\limits_{x\to a_1}f(x)=0.
\displaystyle \text{Also, for }a\neq a_1,a_2,\ldots,a_n,
\displaystyle \lim\limits_{x\to a}f(x)=(a-a_1)(a-a_2)\cdots(a-a_n).
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{Find }\lim\limits_{x\to1^+}\frac{1}{x-1}.
\displaystyle \text{Answer:}
\displaystyle \text{Put }x=1+h,\text{ where }h\to0^+.
\displaystyle \lim\limits_{x\to1^+}\frac{1}{x-1}=\lim\limits_{h\to0^+}\frac{1}{1+h-1}
\displaystyle =\lim\limits_{h\to0^+}\frac{1}{h}=+\infty.
\displaystyle \therefore \lim\limits_{x\to1^+}\frac{1}{x-1}=+\infty.
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{Evaluate the following one-sided limits:}
\displaystyle \text{i) }\lim\limits_{x\to2^+}\frac{x-3}{x^2-4}\qquad \text{ii) }\lim\limits_{x\to2^-}\frac{x-3}{x^2-4}
\displaystyle \text{iii) }\lim\limits_{x\to0^+}\frac{1}{3x}\qquad \text{iv) }\lim\limits_{x\to-8^+}\frac{2x}{x+8}
\displaystyle \text{v) }\lim\limits_{x\to0^+}\frac{2}{x^{\frac15}}\qquad \text{vi) }\lim\limits_{x\to\left(\frac{\pi}{2}\right)^-}\tan x
\displaystyle \text{vii) }\lim\limits_{x\to\left(-\frac{\pi}{2}\right)^+}\sec x\qquad \text{viii) }\lim\limits_{x\to0^-}\frac{x^2-3x+2}{x^3-2x^2}
\displaystyle \text{ix) }\lim\limits_{x\to-2^+}\frac{x^2-1}{2x+4}\qquad \text{x) }\lim\limits_{x\to0^-}(2-\cot x)
\displaystyle \text{xi) }\lim\limits_{x\to0^-}(1+\mathrm{cosec}\,x)
\displaystyle \text{Answer:}
\displaystyle \text{i) Put }x=2+h,\text{ where }h\to0^+.
\displaystyle \lim\limits_{x\to2^+}\frac{x-3}{x^2-4}=\lim\limits_{h\to0^+}\frac{2+h-3}{(2+h)^2-4}
\displaystyle =\lim\limits_{h\to0^+}\frac{h-1}{h(h+4)}=-\infty.
\displaystyle \\

\displaystyle \text{ii) Put }x=2-h,\text{ where }h\to0^+.
\displaystyle \lim\limits_{x\to2^-}\frac{x-3}{x^2-4}=\lim\limits_{h\to0^+}\frac{2-h-3}{(2-h)^2-4}
\displaystyle =\lim\limits_{h\to0^+}\frac{-1-h}{-h(4-h)}=+\infty.
\displaystyle \\

\displaystyle \text{iii) Put }x=0+h,\text{ where }h\to0^+.
\displaystyle \lim\limits_{x\to0^+}\frac{1}{3x}=\lim\limits_{h\to0^+}\frac{1}{3h}=+\infty.
\displaystyle \\

\displaystyle \text{iv) Put }x=-8+h,\text{ where }h\to0^+.
\displaystyle \lim\limits_{x\to-8^+}\frac{2x}{x+8}=\lim\limits_{h\to0^+}\frac{2(-8+h)}{-8+h+8}
\displaystyle =\lim\limits_{h\to0^+}\frac{-16+2h}{h}=-\infty.
\displaystyle \\

\displaystyle \text{v) Put }x=0+h,\text{ where }h\to0^+.
\displaystyle \lim\limits_{x\to0^+}\frac{2}{x^{\frac15}}=\lim\limits_{h\to0^+}\frac{2}{h^{\frac15}}=+\infty.
\displaystyle \\

\displaystyle \text{vi) Put }x=\frac{\pi}{2}-h,\text{ where }h\to0^+.
\displaystyle \lim\limits_{x\to\left(\frac{\pi}{2}\right)^-}\tan x=\lim\limits_{h\to0^+}\tan\left(\frac{\pi}{2}-h\right)
\displaystyle =\lim\limits_{h\to0^+}\cot h=+\infty.
\displaystyle \\

\displaystyle \text{vii) Put }x=-\frac{\pi}{2}+h,\text{ where }h\to0^+.
\displaystyle \lim\limits_{x\to\left(-\frac{\pi}{2}\right)^+}\sec x=\lim\limits_{h\to0^+}\sec\left(-\frac{\pi}{2}+h\right)
\displaystyle =\lim\limits_{h\to0^+}\frac{1}{\sin h}=+\infty.
\displaystyle \\

\displaystyle \text{viii) Put }x=0-h,\text{ where }h\to0^+.
\displaystyle \lim\limits_{x\to0^-}\frac{x^2-3x+2}{x^3-2x^2}=\lim\limits_{h\to0^+}\frac{h^2+3h+2}{-h^3-2h^2}
\displaystyle =\lim\limits_{h\to0^+}\frac{(h+1)(h+2)}{-h^2(h+2)}
\displaystyle =\lim\limits_{h\to0^+}\frac{-(h+1)}{h^2}=-\infty.
\displaystyle \\

\displaystyle \text{ix) Put }x=-2+h,\text{ where }h\to0^+.
\displaystyle \lim\limits_{x\to-2^+}\frac{x^2-1}{2x+4}=\lim\limits_{h\to0^+}\frac{(-2+h)^2-1}{2(-2+h)+4}
\displaystyle =\lim\limits_{h\to0^+}\frac{3-4h+h^2}{2h}=+\infty.
\displaystyle \\

\displaystyle \text{x) Put }x=0-h,\text{ where }h\to0^+.
\displaystyle \lim\limits_{x\to0^-}(2-\cot x)=\lim\limits_{h\to0^+}[2-\cot(-h)]
\displaystyle =\lim\limits_{h\to0^+}(2+\cot h)=+\infty.
\displaystyle \\

\displaystyle \text{xi) Put }x=0-h,\text{ where }h\to0^+.
\displaystyle \lim\limits_{x\to0^-}(1+\mathrm{cosec}\,x)=\lim\limits_{h\to0^+}[1+\mathrm{cosec}(-h)]
\displaystyle =\lim\limits_{h\to0^+}(1-\mathrm{cosec}\,h)=-\infty.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{Show that }\lim\limits_{x\to0}e^{-\frac1x}\text{ does not exist.}
\displaystyle \text{Answer:}
\displaystyle \text{Left-hand limit at }x=0:
\displaystyle \lim\limits_{x\to0^-}e^{-\frac1x}.
\displaystyle \text{Put }x=0-h,\text{ where }h\to0^+.
\displaystyle \lim\limits_{x\to0^-}e^{-\frac1x}=\lim\limits_{h\to0^+}e^{-\frac1{-h}}
\displaystyle =\lim\limits_{h\to0^+}e^{\frac1h}=+\infty.
\displaystyle \text{Right-hand limit at }x=0:
\displaystyle \lim\limits_{x\to0^+}e^{-\frac1x}.
\displaystyle \text{Put }x=0+h,\text{ where }h\to0^+.
\displaystyle \lim\limits_{x\to0^+}e^{-\frac1x}=\lim\limits_{h\to0^+}e^{-\frac1h}
\displaystyle =\lim\limits_{h\to0^+}\frac{1}{e^{\frac1h}}=0.
\displaystyle \therefore \lim\limits_{x\to0^-}e^{-\frac1x}\neq\lim\limits_{x\to0^+}e^{-\frac1x}.
\displaystyle \therefore \lim\limits_{x\to0}e^{-\frac1x}\text{ does not exist.}
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{Find:}
\displaystyle \text{i) }\lim\limits_{x\to2}[x]\hspace{1cm}\text{ii) }\lim\limits_{x\to\frac52}[x]\hspace{1cm}\text{iii) }\lim\limits_{x\to1}[x]
\displaystyle \text{Answer:}
\displaystyle \text{i) }\lim\limits_{x\to2}[x]
\displaystyle \text{Left-hand limit: }\lim\limits_{x\to2^-}[x].
\displaystyle \text{Put }x=2-h,\text{ where }h\to0^+.
\displaystyle \lim\limits_{h\to0^+}[2-h]=1.
\displaystyle \text{Right-hand limit: }\lim\limits_{x\to2^+}[x].
\displaystyle \text{Put }x=2+h,\text{ where }h\to0^+.
\displaystyle \lim\limits_{h\to0^+}[2+h]=2.
\displaystyle \therefore \lim\limits_{x\to2^-}[x]\neq\lim\limits_{x\to2^+}[x].
\displaystyle \therefore \lim\limits_{x\to2}[x]\text{ does not exist.}
\displaystyle \\

\displaystyle \text{ii) }\lim\limits_{x\to\frac52}[x]
\displaystyle \text{Left-hand limit: }\lim\limits_{x\to\left(\frac52\right)^-}[x].
\displaystyle \text{Put }x=\frac52-h,\text{ where }h\to0^+.
\displaystyle \lim\limits_{h\to0^+}\left[\frac52-h\right]=2.
\displaystyle \text{Right-hand limit: }\lim\limits_{x\to\left(\frac52\right)^+}[x].
\displaystyle \text{Put }x=\frac52+h,\text{ where }h\to0^+.
\displaystyle \lim\limits_{h\to0^+}\left[\frac52+h\right]=2.
\displaystyle \therefore \lim\limits_{x\to\left(\frac52\right)^-}[x]=\lim\limits_{x\to\left(\frac52\right)^+}[x].
\displaystyle \therefore \lim\limits_{x\to\frac52}[x]=2.
\displaystyle \\

\displaystyle \text{iii) }\lim\limits_{x\to1}[x]
\displaystyle \text{Left-hand limit: }\lim\limits_{x\to1^-}[x].
\displaystyle \text{Put }x=1-h,\text{ where }h\to0^+.
\displaystyle \lim\limits_{h\to0^+}[1-h]=0.
\displaystyle \text{Right-hand limit: }\lim\limits_{x\to1^+}[x].
\displaystyle \text{Put }x=1+h,\text{ where }h\to0^+.
\displaystyle \lim\limits_{h\to0^+}[1+h]=1.
\displaystyle \therefore \lim\limits_{x\to1^-}[x]\neq\lim\limits_{x\to1^+}[x].
\displaystyle \therefore \lim\limits_{x\to1}[x]\text{ does not exist.}
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{Prove that }\lim\limits_{x\to a^+}[x]=[a]\text{ for all }a\in\mathbb{R}.\text{ Also, prove that }\lim\limits_{x\to1^-}[x]=0.
\displaystyle \text{Answer:}
\displaystyle \text{Right-hand limit at }x=a:
\displaystyle \lim\limits_{x\to a^+}[x].
\displaystyle \text{Put }x=a+h,\text{ where }h\to0^+.
\displaystyle \text{If }a\in\mathbb{Z},\text{ then }[a+h]=a=[a]\text{ for sufficiently small }h>0.
\displaystyle \text{If }a\notin\mathbb{Z},\text{ then }[a+h]=[a]\text{ for sufficiently small }h>0.
\displaystyle \therefore \lim\limits_{x\to a^+}[x]=\lim\limits_{h\to0^+}[a+h]=[a].

\displaystyle \text{Left-hand limit at }x=1:
\displaystyle \lim\limits_{x\to1^-}[x].
\displaystyle \text{Put }x=1-h,\text{ where }h\to0^+.
\displaystyle \therefore \lim\limits_{x\to1^-}[1-h]=0.
\displaystyle \therefore \lim\limits_{x\to1^-}[x]=0.
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{Show that }\lim\limits_{x\to2^-}\frac{x}{[x]}\neq\lim\limits_{x\to2^+}\frac{x}{[x]}.
\displaystyle \text{Answer:}
\displaystyle \text{Left-hand limit at }x=2:
\displaystyle \lim\limits_{x\to2^-}\frac{x}{[x]}.
\displaystyle \text{Put }x=2-h,\text{ where }h\to0^+.
\displaystyle \lim\limits_{h\to0^+}\frac{2-h}{[2-h]}=\lim\limits_{h\to0^+}\frac{2-h}{1}=2.
\displaystyle \text{Right-hand limit at }x=2:
\displaystyle \lim\limits_{x\to2^+}\frac{x}{[x]}.
\displaystyle \text{Put }x=2+h,\text{ where }h\to0^+.
\displaystyle \lim\limits_{h\to0^+}\frac{2+h}{[2+h]}=\lim\limits_{h\to0^+}\frac{2+h}{2}=1.
\displaystyle \therefore \lim\limits_{x\to2^-}\frac{x}{[x]}\neq\lim\limits_{x\to2^+}\frac{x}{[x]}.
\displaystyle \therefore \lim\limits_{x\to2}\frac{x}{[x]}\text{ does not exist.}
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{Find }\lim\limits_{x\to3^+}\frac{x}{[x]}. \text{ Is it equal to }\lim\limits_{x\to3^-}\frac{x}{[x]}?
\displaystyle \text{Answer:}
\displaystyle \text{Left-hand limit at }x=3:
\displaystyle \lim\limits_{x\to3^-}\frac{x}{[x]}.
\displaystyle \text{Put }x=3-h,\text{ where }h\to0^+.
\displaystyle \lim\limits_{h\to0^+}\frac{3-h}{[3-h]}=\lim\limits_{h\to0^+}\frac{3-h}{2}=\frac32.
\displaystyle \text{Right-hand limit at }x=3:
\displaystyle \lim\limits_{x\to3^+}\frac{x}{[x]}.
\displaystyle \text{Put }x=3+h,\text{ where }h\to0^+.
\displaystyle \lim\limits_{h\to0^+}\frac{3+h}{[3+h]}=\lim\limits_{h\to0^+}\frac{3+h}{3}=1.
\displaystyle \therefore \lim\limits_{x\to3^+}\frac{x}{[x]}=1.
\displaystyle \text{Also, }\lim\limits_{x\to3^-}\frac{x}{[x]}=\frac32\neq1.
\displaystyle \therefore \lim\limits_{x\to3^+}\frac{x}{[x]}\neq\lim\limits_{x\to3^-}\frac{x}{[x]}.
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{Find }\lim\limits_{x\to\frac52}[x].
\displaystyle \text{Answer:}
\displaystyle \text{Left-hand limit at }x=\frac52:
\displaystyle \lim\limits_{x\to\left(\frac52\right)^-}[x].
\displaystyle \text{Put }x=\frac52-h,\text{ where }h\to0^+.
\displaystyle \lim\limits_{h\to0^+}\left[\frac52-h\right]=2.
\displaystyle \text{Right-hand limit at }x=\frac52:
\displaystyle \lim\limits_{x\to\left(\frac52\right)^+}[x].
\displaystyle \text{Put }x=\frac52+h,\text{ where }h\to0^+.
\displaystyle \lim\limits_{h\to0^+}\left[\frac52+h\right]=2.
\displaystyle \therefore \lim\limits_{x\to\left(\frac52\right)^-}[x]=\lim\limits_{x\to\left(\frac52\right)^+}[x].
\displaystyle \therefore \lim\limits_{x\to\frac52}[x]=2.
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{Evaluate }\lim\limits_{x\to2}f(x)\text{ if it exists, where }f(x)=\left\{\begin{array}{ll}x-[x],&x<2\\4,&x=2\\3x-5,&x>2.\end{array}\right.
\displaystyle \text{Answer:}
\displaystyle \text{Left-hand limit at }x=2:
\displaystyle \lim\limits_{x\to2^-}(x-[x]).
\displaystyle \text{Put }x=2-h,\text{ where }h\to0^+.
\displaystyle \lim\limits_{h\to0^+}\big((2-h)-[2-h]\big)=\lim\limits_{h\to0^+}\big((2-h)-1\big)=1.
\displaystyle \text{Right-hand limit at }x=2:
\displaystyle \lim\limits_{x\to2^+}(3x-5).
\displaystyle \text{Put }x=2+h,\text{ where }h\to0^+.
\displaystyle \lim\limits_{h\to0^+}\big(3(2+h)-5\big)=1.
\displaystyle \therefore \lim\limits_{x\to2^-}f(x)=\lim\limits_{x\to2^+}f(x)=1.
\displaystyle \therefore \lim\limits_{x\to2}f(x)=1.
\displaystyle \text{Note that }f(2)=4,\text{ but it does not affect the value of the limit.}
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{Show that }\lim\limits_{x\to0}\sin\frac1x\text{ does not exist.}
\displaystyle \text{Answer:}
\displaystyle \text{Consider the sequences }x_n=\frac{1}{\frac{\pi}{2}+2n\pi}\text{ and }y_n=\frac{1}{\frac{3\pi}{2}+2n\pi}.
\displaystyle \text{As }n\to\infty,\ x_n\to0^+\text{ and }y_n\to0^+.
\displaystyle \sin\frac1{x_n}=\sin\left(\frac{\pi}{2}+2n\pi\right)=1.
\displaystyle \sin\frac1{y_n}=\sin\left(\frac{3\pi}{2}+2n\pi\right)=-1.
\displaystyle \text{Thus, along two sequences approaching }0^+,\ \sin\frac1x\text{ has different limiting values.}
\displaystyle \therefore \lim\limits_{x\to0^+}\sin\frac1x\text{ does not exist.}
\displaystyle \therefore \lim\limits_{x\to0}\sin\frac1x\text{ does not exist.}
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{Let }f(x)=\left\{\begin{array}{ll}\dfrac{k\cos x}{\pi-2x},&x\neq\dfrac{\pi}{2}\\[4pt]3,&x=\dfrac{\pi}{2}\end{array}\right.\text{ and if }\lim\limits_{x\to\frac{\pi}{2}}f(x)=f\!\left(\frac{\pi}{2}\right),\text{ find }k.
\displaystyle \text{Answer:}
\displaystyle \text{Since }\lim\limits_{x\to\frac{\pi}{2}}f(x)=f\!\left(\frac{\pi}{2}\right),\text{ we have}
\displaystyle \lim\limits_{x\to\frac{\pi}{2}}\frac{k\cos x}{\pi-2x}=3.
\displaystyle \text{Applying L'Hopital's Rule,}
\displaystyle \lim\limits_{x\to\frac{\pi}{2}}\frac{k\cos x}{\pi-2x}=\lim\limits_{x\to\frac{\pi}{2}}\frac{-k\sin x}{-2}=\frac{k}{2}.
\displaystyle \therefore \frac{k}{2}=3.
\displaystyle \therefore k=6.
\displaystyle \\

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