Class 11: Limits – Exercise 29.1 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

$\displaystyle \text{Question 1: Show that } \lim \limits_{x \to 0} \frac{x}{|x|} \text{ does not exist.}$

Answer:

$\displaystyle \text{LHL of } \lim \limits_{x \to 0} \ \frac{x}{|x|}:$ $\displaystyle \lim \limits_{x \to 0^-} \frac{x}{|x|}$

$\displaystyle \text{Let } x = 0-h, \text{ where } h \rightarrow 0$

$\displaystyle \Rightarrow \lim \limits_{h \to 0} \frac{0-h}{|0-h|} = \lim \limits_{h \to 0} \frac{-h}{h} = -1$

$\displaystyle \text{RHL of } \lim \limits_{x \to 0} \ \frac{x}{|x|}:$ $\displaystyle \lim \limits_{x \to 0^+} \frac{x}{|x|}$

$\displaystyle \text{Let } x = 0+h, \text{ where } h \rightarrow 0$

$\displaystyle \Rightarrow \lim \limits_{h \to 0} \frac{0+h}{|0+h|} = \lim \limits_{h \to 0} \frac{h}{h} = 1$

$\displaystyle \text{Clearly, } \lim \limits_{x \to 0^-} \ \frac{x}{|x|} \neq \lim \limits_{x \to 0^+} \frac{x}{|x|} .$

$\displaystyle \text{Hence, } \lim \limits_{x \to 0} \frac{x}{|x|} \text{ does not exist.}$

$\\$

$\displaystyle \text{Question 2: Find } k \text{ so that} \lim \limits_{x \to 2} f(x) \text{ may exist, where } \\ \\ f(x) = \Bigg\{ \begin{array}{lll} 2x+3, & x \leq 2 \\ x+k, & x >2 \end{array}$

Answer:

$\displaystyle \text{LHL of } f(x) \text{ at } x = 2: \lim \limits_{x \to 2^-} f(x) = \lim \limits_{x \to 2^-} 2x+3$

$\displaystyle \text{Let } x = 2-h, \text{ where } h \rightarrow 0$

$\displaystyle \Rightarrow \lim \limits_{h \to 0} (2(2-h)+3) = 7$

$\displaystyle \text{RHL of } f(x) \text{ at } x = 2: \lim \limits_{x \to 2^+} f(x) = \lim \limits_{x \to 2^+} x+k$

$\displaystyle \text{Let } x = 2+h, \text{ where } h \rightarrow 0$

$\displaystyle \Rightarrow \lim \limits_{h \to 0} (2+h+k) = 2+k$

$\displaystyle \text{Now, } \lim \limits_{x \to 2^-} f(x) \text{ exists if the LHL = RHL }$

$\displaystyle \Rightarrow 7 = 2+k \Rightarrow k = 5$

$\\$

$\displaystyle \text{Question 3: Show that } \lim \limits_{x \to 0} \frac{1}{x} \text{ does not exist.}$

Answer:

$\displaystyle \text{LHL of } \lim \limits_{x \to 0} \frac{1}{x}: \lim \limits_{x \to 0^-} \frac{1}{x}$

$\displaystyle \text{Let } x = 0-h, \text{ where } h \rightarrow 0$

$\displaystyle \Rightarrow \lim \limits_{h \to 0} \frac{1}{0-h}= - \infty$

$\displaystyle \text{LHL of } \lim \limits_{x \to 0} \frac{1}{x}: \lim \limits_{x \to 0^+} \frac{1}{x}$

$\displaystyle \text{Let } x = 0+h, \text{ where } h \rightarrow 0$

$\displaystyle \Rightarrow \lim \limits_{h \to 0} \frac{1}{0+h}= \infty$

$\displaystyle \text{Clearly, } \lim \limits_{x \to 0^-} \ \frac{1}{x} \neq \lim \limits_{x \to 0^+} \frac{1}{x} .$

$\displaystyle \text{Hence, } \lim \limits_{x \to 0} \ \frac{1}{x} \text{ does not exist.}$

$\\$

$\displaystyle \text{Question 4: Let } f(x) \text{ be a function defined by } f(x) = \Bigg\{ \begin{array}{ll} \frac{3x}{|x|+2x} , & x \neq 0 \\ 0, & x =0 \end{array}$

$\displaystyle \text{Show that } \lim \limits_{x \to 0} f(x) \text{ does not exist.}$

Answer:

$\displaystyle \text{LHL of } f(x) \text{ at } x = 0: \lim \limits_{x \to 0^-} f(x) = \lim \limits_{x \to 0^-} \frac{3x}{|x|+2x}$

$\displaystyle \text{Let } x = 0-h, \text{ where } h \rightarrow 0$

$\displaystyle \Rightarrow \lim \limits_{h \to 0} \frac{3(-h)}{|-h|+2(-h)} = \frac{-3h}{h - 2h} = \frac{-3h}{-h} = 3$

$\displaystyle \text{RHL of } f(x) \text{ at } x = 0: \lim \limits_{x \to 0^+} f(x) = \lim \limits_{x \to 0^+} \frac{3x}{|x|+2x}$

$\displaystyle \text{Let } x = 0+h, \text{ where } h \rightarrow 0$

$\displaystyle \Rightarrow \lim \limits_{h \to 0} \frac{3(h)}{|h|+2(h)} = \frac{3h}{h + 2h} = \frac{3h}{3h} = 1$

$\displaystyle \text{Clearly, } \lim \limits_{x \to 0^-} \ \frac{3x}{|x|+2x} \neq \lim \limits_{x \to 0^+} \frac{3x}{|x|+2x} .$

$\displaystyle \text{Hence, } \lim \limits_{x \to 0} \ f(x) \text{ does not exist.}$

$\\$

$\displaystyle \text{Question 5: Let } f(x) = \Bigg\{ \begin{array}{ll} x+1 , & \text{ if } x > 0 \\ x-1, & \text{ if } x <0 \end{array}$

$\displaystyle \text{Prove that } \lim \limits_{x \to 0} f(x) \text{ does not exist.}$

Answer:

$\displaystyle \text{LHL of } f(x) \text{ at } x = 0: \lim \limits_{x \to 0^-} f(x) = \lim \limits_{x \to 0^-} (x-1)$

$\displaystyle \text{Let } x = 0-h, \text{ where } h \rightarrow 0$

$\displaystyle \Rightarrow \lim \limits_{h \to 0} (0-h-1)= -1$

$\displaystyle \text{RHL of } f(x) \text{ at } x = 0: \lim \limits_{x \to 0^+} f(x) = \lim \limits_{x \to 0^+} (x+1)$

$\displaystyle \text{Let } x = 0+h, \text{ where } h \rightarrow 0$

$\displaystyle \Rightarrow \lim \limits_{h \to 0} (0+h+1) = 1$

$\displaystyle \text{Clearly, } \lim \limits_{x \to 0^-} \ f(x) \neq \lim \limits_{x \to 0^+} f(x) .$

$\displaystyle \text{Hence, } \lim \limits_{x \to 0} \ f(x) \text{ does not exist.}$

$\\$

$\displaystyle \text{Question 6: Let } f(x) = \Bigg\{ \begin{array}{ll} x+5 , & \text{ if } x > 0 \\ x-4, & \text{ if } x <0 \end{array}$

$\displaystyle \text{Prove that } \lim \limits_{x \to 0} f(x) \text{ does not exist.}$

Answer:

$\displaystyle \text{LHL of } f(x) \text{ at } x = 0: \lim \limits_{x \to 0^-} f(x) = \lim \limits_{x \to 0^-} (x-4)$

$\displaystyle \text{Let } x = 0-h, \text{ where } h \rightarrow 0$

$\displaystyle \Rightarrow \lim \limits_{h \to 0} (0-h-4)= -4$

$\displaystyle \text{RHL of } f(x) \text{ at } x = 0: \lim \limits_{x \to 0^+} f(x) = \lim \limits_{x \to 0^+} (x+5)$

$\displaystyle \text{Let } x = 0+h, \text{ where } h \rightarrow 0$

$\displaystyle \Rightarrow \lim \limits_{h \to 0} (0+h+5) = 5$

$\displaystyle \text{Clearly, } \lim \limits_{x \to 0^-} \ f(x) \neq \lim \limits_{x \to 0^+} f(x) .$

$\displaystyle \text{Hence, } \lim \limits_{x \to 0} \ f(x) \text{ does not exist.}$

$\\$

$\displaystyle \text{Question 7: Find } \lim \limits_{x \to 3} f(x) \text{, where } f(x) = \Bigg\{ \begin{array}{ll} 4 , & \text{ if } x > 3 \\ x+1, & \text{ if } x <3 \end{array}$

Answer:

$\displaystyle \text{LHL of } f(x) \text{ at } x = 3: \lim \limits_{x \to 3^-} f(x) = \lim \limits_{x \to 3^-} (x+1)$

$\displaystyle \text{Let } x = 3-h, \text{ where } h \rightarrow 0$

$\displaystyle \Rightarrow \lim \limits_{h \to 3} (3-h+1)= 4$

$\displaystyle \text{RHL of } f(x) \text{ at } x = 3: \lim \limits_{x \to 3^+} f(x) = \lim \limits_{x \to 3^+} 4$

$\displaystyle \text{Let } x = 3+h, \text{ where } h \rightarrow 0$

$\displaystyle \Rightarrow \lim \limits_{h \to 3} 4 = 4$

$\displaystyle \text{Clearly, } \lim \limits_{x \to 3^-} \ f(x) = \lim \limits_{x \to 3^+} f(x) .$

$\displaystyle \text{Hence, } \lim \limits_{x \to 3} \ f(x) = 4$

$\\$

$\displaystyle \text{Question 8: If } f(x) = \Bigg\{ \begin{array}{ll} 2x+3 , & x \leq 0 \\ 3(x+1), & x >0 \end{array}$

$\displaystyle \text{Find } \lim \limits_{x \to 0} f(x) \text{ and } \lim \limits_{x \to 1} f(x).$

Answer:

$\displaystyle \text{LHL of } f(x) \text{ at } x = 0: \lim \limits_{x \to 0^-} f(x) = \lim \limits_{x \to 0^-} (2x+3)$

$\displaystyle \text{Let } x = 0-h, \text{ where } h \rightarrow 0$

$\displaystyle \Rightarrow \lim \limits_{h \to 0} (2(0-h)+3)= \lim \limits_{h \to 0} (-2h+3) = 3$

$\displaystyle \text{RHL of } f(x) \text{ at } x = 0: \lim \limits_{x \to 0^+} f(x) = \lim \limits_{x \to 0^+} 3(x+1)$

$\displaystyle \text{Let } x = 0+h, \text{ where } h \rightarrow 0$

$\displaystyle \Rightarrow \lim \limits_{h \to 0} 3((0+h)+1)= \lim \limits_{h \to 0} (3h+3) = 3$

$\displaystyle \text{Clearly, } \lim \limits_{x \to 0^-} \ f(x) = \lim \limits_{x \to 0^+} f(x) .$

$\displaystyle \text{Hence, } \lim \limits_{x \to 0} \ f(x) = 3$

$\\$

$\displaystyle \text{LHL of } f(x) \text{ at } x = 1: \lim \limits_{x \to 1^-} f(x) = \lim \limits_{x \to 1^-} 3(x+1)$

$\displaystyle \text{Let } x = 1-h, \text{ where } h \rightarrow 0$

$\displaystyle \Rightarrow \lim \limits_{h \to 1} 3(1-h+1)= \lim \limits_{h \to 1} (2-h) = 6$

$\displaystyle \text{RHL of } f(x) \text{ at } x = 1: \lim \limits_{x \to 1^+} f(x) = \lim \limits_{x \to 1^+} 3(x+1)$

$\displaystyle \text{Let } x = 1+h, \text{ where } h \rightarrow 0$

$\displaystyle \Rightarrow \lim \limits_{h \to 1} 3(1+h+1)= \lim \limits_{h \to 1} (2+h) = 6$

$\displaystyle \text{Clearly, } \lim \limits_{x \to 1^-} \ f(x) = \lim \limits_{x \to 1^+} f(x) .$

$\displaystyle \text{Hence, } \lim \limits_{x \to 1} \ f(x) = 6$

$\\$

$\displaystyle \text{Question 9: Find } \lim \limits_{x \to 1} f(x), \text{ if } f(x) = \Bigg\{ \begin{array}{ll} x^2-1 , & x \leq 1 \\ -x^2-1, & x >1 \end{array}$

Answer:

$\displaystyle \text{LHL of } f(x) \text{ at } x = 1: \lim \limits_{x \to 1^-} f(x) = \lim \limits_{x \to 1^-} (x^2-1)$

$\displaystyle \text{Let } x = 1-h, \text{ where } h \rightarrow 0$

$\displaystyle \Rightarrow \lim \limits_{h \to 1} (1-h)^2-1= \lim \limits_{h \to 1} = 0$

$\displaystyle \text{RHL of } f(x) \text{ at } x = 1: \lim \limits_{x \to 1^+} f(x) = \lim \limits_{x \to 1^+} -x^2-1$

$\displaystyle \text{Let } x = 1+h, \text{ where } h \rightarrow 0$

$\displaystyle \Rightarrow \lim \limits_{h \to 1} -(1+h)^2-1= -2$

$\displaystyle \text{Clearly, } \lim \limits_{x \to 1^-} \ f(x) \neq \lim \limits_{x \to 1^+} f(x) .$

$\displaystyle \text{Hence, } \lim \limits_{x \to 1} \text{ does not exist.}$

$\\$

$\displaystyle \text{Question 10: Evaluate } \lim \limits_{x \to 0} f(x) \text{ where } f(x) = \Bigg\{ \begin{array}{ll} \frac{|x|}{x} , & x \neq 0 \\ 0, & x =0 \end{array}$

Answer:

$\displaystyle \text{LHL of } f(x) \text{ at } x = 0: \lim \limits_{x \to 0^-} f(x) = \lim \limits_{x \to 0^-} \frac{|x|}{x}$

$\displaystyle \text{Let } x = 0-h, \text{ where } h \rightarrow 0$

$\displaystyle \Rightarrow \lim \limits_{h \to 0} \frac{|0-h|}{0-h}= \lim \limits_{h \to 0} \frac{h}{-h} = -1$

$\displaystyle \text{RHL of } f(x) \text{ at } x = 0: \lim \limits_{x \to 0^+} f(x) = \lim \limits_{x \to 0^+} \frac{|x|}{x}$

$\displaystyle \text{Let } x = 0+h, \text{ where } h \rightarrow 0$

$\displaystyle \Rightarrow \lim \limits_{h \to 0} \frac{|0+h|}{0+h}= \lim \limits_{h \to 0} \frac{h}{h} = 1$

$\displaystyle \text{Clearly, } \lim \limits_{x \to 0^-} \ f(x) \neq \lim \limits_{x \to 0^+} f(x) .$

$\displaystyle \text{Hence, } \lim \limits_{x \to 0} \ f(x) \text{ does not exist.}$

$\\$

$\displaystyle \text{Question 11: Let } a_1, a_2, \cdots , a_n \text{ be fixed real numbers such that } \\ \\ f(x) = (x-a_1)(x-a_2) \cdots (x-a_n). \\ \\ \text{What is } \lim \limits_{x \to a_1} f(x)? \text{ For a } \neq a_1, a_2, \cdots , a_n \text{ compute }\lim \limits_{x \to a} f(x).$

Answer:

$\displaystyle \text{LHL of } f(x) \text{ at } x = a_1: \lim \limits_{x \to {a_1}^-} f(x) = \lim \limits_{x \to {a_1}^-} (x-a_1)(x-a_2) \cdots (x-a_n)$

$\displaystyle \text{Let } x = a_1-h, \text{ where } h \rightarrow 0$

$\displaystyle \Rightarrow \lim \limits_{h \to 0} (a_1-h-a_1)(a_1-h-a_2) \cdots (a_1-h-a_n)= 0$

$\displaystyle \text{RHL of } f(x) \text{ at } x = a_1: \lim \limits_{x \to {a_1}^+} f(x) = \lim \limits_{x \to {a_1}^+} (x-a_1)(x-a_2) \cdots (x-a_n)$

$\displaystyle \text{Let } x = a_1+h, \text{ where } h \rightarrow 0$

$\displaystyle \Rightarrow \lim \limits_{h \to 0} (a_1+h-a_1)(a_1+h-a_2) \cdots (a_1+h-a_n)= 0$

$\displaystyle \text{Clearly, } \lim \limits_{x \to a_1^-} \ f(x) = \lim \limits_{x \to a_1^+} f(x) .$

$\displaystyle \text{Hence, } \lim \limits_{x \to a_1} \ f(x) = 0$

$\displaystyle \text{Similarly, } \lim \limits_{x \to a} \ f(x) = (a-a_1)(a-a_2) \cdots (a-a_n)$

$\\$

$\displaystyle \text{Question 12: Find } \lim \limits_{x \to {1^+}} \frac{1}{x-1}$

Answer:

$\displaystyle \text{RHL of } f(x) \text{ at } x = 1: \lim \limits_{x \to 1^+} f(x) = \lim \limits_{x \to 1^+} \frac{1}{x-1}$

$\displaystyle \text{Let } x = 1+h, \text{ where } h \rightarrow 0$

$\displaystyle \Rightarrow \lim \limits_{h \to 0} \frac{1}{1+h-1}= \lim \limits_{h \to 0} \ \frac{1}{h} = \infty$

$\\$

Question 13: Evaluate the following one sides limits:

$\displaystyle \text{ i) } \lim \limits_{x \to {2^+}} \frac{x-3}{x^2-4}$ $\displaystyle \text{ ii) } \lim \limits_{x \to {2^-}} \frac{x-3}{x^2-4}$ $\displaystyle \text{ iii) } \lim \limits_{x \to {0^+}} \frac{1}{3x}$

$\displaystyle \text{ iv) } \lim \limits_{x \to {{-8}^+}} \frac{2x}{x+8}$ $\displaystyle \text{ v) } \lim \limits_{x \to {0^+}} \frac{2}{x^{\frac{1}{5}}}$ $\displaystyle \text{ vi) } \lim \limits_{x \to {\frac{\pi^-}{2}}} \tan x$

$\displaystyle \text{ vii) } \lim \limits_{x \to {-\frac{\pi}{2^+}}} \sec x$ $\displaystyle \text{ viii) } \lim \limits_{x \to {0^-}} \frac{x^2-3x+2}{x^3-2x^2}$ $\displaystyle \text{ ix) } \lim \limits_{x \to {-2^+}} \frac{x^2-1}{2x+4}$

$\displaystyle \text{ x) } \lim \limits_{x \to {0^-}} (2-\cot x)$ $\displaystyle \text{ xi) } \lim \limits_{x \to {0^-}} (1+ \mathrm{cosec} x)$

Answer:

$\displaystyle \text{ i) } \text{Given: } \lim \limits_{x \to 2^+} \frac{x-3}{x^2-4}$

$\displaystyle \text{Let } x = 2+h, \text{ where } h \rightarrow 0$

$\displaystyle \Rightarrow \lim \limits_{h \to 0} \frac{2+h-3}{(2+h)^2-4} = \frac{-1+h}{4 + h^2 + 4h - 4} = - \infty$

$\displaystyle \text{ ii) } \text{Given: } \lim \limits_{x \to {2^-}} \frac{x-3}{x^2-4}$

$\displaystyle \text{Let } x = 2-h, \text{ where } h \rightarrow 0$

$\displaystyle \Rightarrow \lim \limits_{h \to 0} \frac{2-h-3}{(2-h)^2-4} = \frac{-1-h}{4 + h^2 - 4h - 4} = \frac{-1-h}{-h(4-h)} = \infty$

$\displaystyle \text{ iii) } \text{Given: } \lim \limits_{x \to 0^+} \frac{1}{3x}$

$\displaystyle \text{Let } x = 0+h, \text{ where } h \rightarrow 0$

$\displaystyle \Rightarrow \lim \limits_{h \to 0} \frac{1}{3(0+h)} = \frac{1}{3h} = \infty$

$\displaystyle \text{ iv) } \text{Given: } \lim \limits_{x \to 8^+} \frac{2x}{x+8}$

$\displaystyle \text{Let } x = -8+h, \text{ where } h \rightarrow 0$

$\displaystyle \Rightarrow \lim \limits_{h \to 0} \frac{2(-8+h)}{-8+h+8} = \frac{-16+2h}{h} = -\infty$

$\displaystyle \text{ v) } \text{Given: } \lim \limits_{x \to 0^+} \frac{2}{x^{\frac{1}{5}}}$

$\displaystyle \text{Let } x = 0+h, \text{ where } h \rightarrow 0$

$\displaystyle \Rightarrow \lim \limits_{h \to 0} \frac{2}{(0+h)^{\frac{1}{5}}} = \frac{2}{h^{\frac{1}{5}}} = \infty$

$\displaystyle \text{ vi) } \text{Given: } \lim \limits_{x \to {\frac{\pi}{2}}^-} \tan x$

$\displaystyle \text{Let } x = \frac{\pi}{2} - h, \text{ where } h \rightarrow 0$

$\displaystyle \Rightarrow \lim \limits_{h \to 0} \tan (\frac{\pi}{2} - h) = \lim \limits_{h \to 0} \cot h = \infty$

$\displaystyle \text{ vii) } \text{Given: } \lim \limits_{x \to {-\frac{\pi}{2}}^+} \tan x$

$\displaystyle \text{Let } x = -\frac{\pi}{2} + h, \text{ where } h \rightarrow 0$

$\displaystyle \Rightarrow \lim \limits_{h \to 0} \sec ( -\frac{\pi}{2} + h) = \lim \limits_{h \to 0} \sec ( \frac{\pi}{2} - h) = \lim \limits_{h \to 0} \mathrm{cosec} h= \infty$

$\displaystyle \text{ viii) } \text{Given: } \lim \limits_{x \to 0^-} \frac{x^2-3x+2}{x^3-2x^2}$

$\displaystyle \text{Let } x = 0-h, \text{ where } h \rightarrow 0$

$\displaystyle \Rightarrow \lim \limits_{h \to 0} \frac{(0-h)^2-3(0-h)+2}{(0-h)^3-2(0-h)^2} \\ \\ = \frac{h^2+3h+2}{-h^3-2h^2} = \frac{(h+2)(h+1)}{-h^2(h+2)} = \frac{h+1}{-h^2} =-\infty$

$\displaystyle \text{ ix) } \text{Given: } \lim \limits_{x \to {-2^+}} \frac{x^2-1}{2x+4}$

$\displaystyle \text{Let } x = -2+h, \text{ where } h \rightarrow 0$

$\displaystyle \Rightarrow \lim \limits_{h \to 0} \frac{(-2+h)^2-1}{2(-2+h)+4} = \frac{4+h^2-4h-1}{-4+2h+4} = \frac{3+h^2-3h}{2h} = \infty$

$\displaystyle \text{ x) } \text{Given: } \lim \limits_{x \to 0^-} (2-\cot x)$

$\displaystyle \text{Let } x = 0-h, \text{ where } h \rightarrow 0$

$\displaystyle \Rightarrow \lim \limits_{h \to 0} (2-\cot (0-h)) = \lim \limits_{h \to 0} (2+\cot h) = \infty$

$\displaystyle \text{ xi) } \text{Given: } \lim \limits_{x \to 0^-} (1+ \mathrm{cosec} x)$

$\displaystyle \text{Let } x = 0-h, \text{ where } h \rightarrow 0$

$\displaystyle \Rightarrow \lim \limits_{h \to 0} (1+ \mathrm{cosec} (0-h)) = \lim \limits_{h \to 0} (1- \mathrm{cosec} h) = -\infty$

$\\$

$\displaystyle \text{Question 14: Show that } \lim \limits_{x \to 0} e^{\frac{-1}{x}} \text{ does not exist.}$

Answer:

$\displaystyle \text{LHL of } f(x) \text{ at } x = 0: \lim \limits_{x \to 0^-} f(x) = \lim \limits_{x \to 0^-} e^{\frac{-1}{x}}$

$\displaystyle \text{Let } x = 0-h, \text{ where } h \rightarrow 0$

$\displaystyle \Rightarrow \lim \limits_{h \to 0} e^{\frac{-1}{0-h}}= \lim \limits_{h \to 0} e^{\frac{1}{h}} = e^{\infty} = \infty$

$\displaystyle \text{RHL of } f(x) \text{ at } x = 0: \lim \limits_{x \to 0^+} f(x) = \lim \limits_{x \to 0^+} e^{\frac{-1}{x}}$

$\displaystyle \text{Let } x = 0+h, \text{ where } h \rightarrow 0$

$\displaystyle \Rightarrow \lim \limits_{h \to 0} e^{\frac{-1}{0+h}}= \lim \limits_{h \to 0} e^{\frac{-1}{h}} = \lim \limits_{h \to 0} \frac{1}{e^{\frac{1}{h}}} = \frac{1}{\infty} = 0$

$\displaystyle \text{Clearly, } \lim \limits_{x \to 0^-} \ e^{\frac{-1}{x}} \neq \lim \limits_{x \to 0^+} e^{\frac{-1}{x}} .$

$\displaystyle \text{Hence, } \lim \limits_{x \to 0} \ e^{\frac{-1}{x}} \text{ does not exist.}$

$\\$

Question 15: Find:

$\displaystyle \text{ i) } \lim \limits_{x \to 2} [x] \hspace{1.0cm} \text{ ii) } \lim \limits_{x \to \frac{5}{2} } [x] \hspace{1.0cm} \text{ iii) } \lim \limits_{x \to 1} [x]$

Answer:

$\displaystyle \text{ i) } \lim \limits_{x \to 2} [x]$

$\displaystyle \text{LHL of } \lim \limits_{x \to 2} [x] : \lim \limits_{x \to 2^-} [x]$

$\displaystyle \text{Let } x = 2-h, \text{ where } h \rightarrow 0$

$\displaystyle \Rightarrow \lim \limits_{h \to 0} [ 2-h] = 1$

$\displaystyle \text{RHL of } \lim \limits_{x \to 2} [x] : \lim \limits_{x \to 2^+} [x]$

$\displaystyle \text{Let } x = 2+h, \text{ where } h \rightarrow 0$

$\displaystyle \Rightarrow \lim \limits_{h \to 0} [ 2+h] = 2$

$\displaystyle \text{Clearly, } \lim \limits_{x \to 2^-} [x] \neq \lim \limits_{x \to 2^+} [x] .$

$\displaystyle \text{Hence, } \lim \limits_{x \to 2} \ [x] \text{ does not exist.}$

$\\$

$\displaystyle \text{ iii) } \lim \limits_{x \to \frac{5}{2} } [x]$

$\displaystyle \text{LHL of } \lim \limits_{x \to \frac{5}{2}} [x] : \lim \limits_{x \to {\frac{5}{2}}^-} [x]$

$\displaystyle \text{Let } x = \frac{5}{2}-h, \text{ where } h \rightarrow 0$

$\displaystyle \Rightarrow \lim \limits_{h \to 0} [ \frac{5}{2} - h ] = 2$

$\displaystyle \text{RHL of } \lim \limits_{x \to \frac{5}{2}} [x] : \lim \limits_{x \to {\frac{5}{2}}^+} [x]$

$\displaystyle \text{Let } x = \frac{5}{2}+h, \text{ where } h \rightarrow 0$

$\displaystyle \Rightarrow \lim \limits_{h \to 0} [ \frac{5}{2} + h ] = 2$

$\displaystyle \text{Clearly, } \lim \limits_{x \to 2^-} [x] = \lim \limits_{x \to 2^+} [x] .$

$\displaystyle \text{Hence, } \lim \limits_{x \to \frac{5}{2} } [x] = 2$

$\\$

$\displaystyle \text{ ii) } \lim \limits_{x \to 1} [x]$

$\displaystyle \text{LHL of } \lim \limits_{x \to 1} [x] : \lim \limits_{x \to 1^-} [x]$

$\displaystyle \text{Let } x = 1-h, \text{ where } h \rightarrow 0$

$\displaystyle \Rightarrow \lim \limits_{h \to 0} [ 1-h] = 0$

$\displaystyle \text{RHL of } \lim \limits_{x \to 1} [x] : \lim \limits_{x \to 1^+} [x]$

$\displaystyle \text{Let } x = 1+h, \text{ where } h \rightarrow 0$

$\displaystyle \Rightarrow \lim \limits_{h \to 0} [ 1+h] = 1$

$\displaystyle \text{Clearly, } \lim \limits_{x \to 1^-} [x] \neq \lim \limits_{x \to 1^+} [x] .$

$\displaystyle \text{Hence, } \lim \limits_{x \to 1} \ [x] \text{ does not exist.}$

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$\displaystyle \text{Question 16: Prove that} \lim \limits_{x \to a^+} [x] \text{ for all } a \in R. \text{ Also, prove that} \lim \limits_{x \to 1^-} [x] = 0.$

Answer:

$\displaystyle \text{RHL of } \lim \limits_{x \to a} [x] : \lim \limits_{x \to a^+} [x]$

$\displaystyle \text{Let } x = a+h, \text{ where } h \rightarrow 0$

$\displaystyle \Rightarrow \lim \limits_{h \to 0} [ a+h] = a$

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$\displaystyle \text{LHL of } \lim \limits_{x \to 1} [x] : \lim \limits_{x \to 1^-} [x]$

$\displaystyle \text{Let } x = 1-h, \text{ where } h \rightarrow 0$

$\displaystyle \Rightarrow \lim \limits_{h \to 0} [ 1-h] = 0$

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$\displaystyle \text{Question 17: Show that} \lim \limits_{x \to 2^-} \frac{x}{[x]} \neq \lim \limits_{x \to 2^+} \frac{x}{[x]} .$

Answer:

$\displaystyle \text{LHL of } \lim \limits_{x \to 2} \ \frac{x}{[x]}: \lim \limits_{x \to 2^-} \frac{x}{[x]}$

$\displaystyle \text{Let } x = 2-h, \text{ where } h \rightarrow 0$

$\displaystyle \Rightarrow \lim \limits_{h \to 0} \frac{2-h}{[2-h] } = \lim \limits_{h \to 0} \frac{2}{1} = 2$

$\displaystyle \text{RHL of } \lim \limits_{x \to 2} \ \frac{x}{[x]}: \lim \limits_{x \to 2^+} \frac{x}{|x|}$

$\displaystyle \text{Let } x = 2+h, \text{ where } h \rightarrow 0$

$\displaystyle \Rightarrow \lim \limits_{h \to 0} \frac{2+h}{[2+h]} = \lim \limits_{h \to 0} \frac{2}{2} = 1$

$\displaystyle \text{Clearly, } \lim \limits_{x \to 2^-} \frac{x}{[x]} \neq \lim \limits_{x \to 2^+} \frac{x}{|x|} .$

$\displaystyle \text{Hence, } \lim \limits_{x \to 2} \frac{x}{|x|} \text{ does not exist.}$

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$\displaystyle \text{Question 18: Find } \lim \limits_{x \to 3^+} \frac{x}{[x]}. \text{ Is it equal to } \lim \limits_{x \to 3^-} \frac{x}{[x]} .$

Answer:

$\displaystyle \text{LHL of } \lim \limits_{x \to 3} \ \frac{x}{[x]}: \lim \limits_{x \to 3^-} \frac{x}{[x]}$

$\displaystyle \text{Let } x = 3-h, \text{ where } h \rightarrow 0$

$\displaystyle \Rightarrow \lim \limits_{h \to 0} \frac{3-h}{[3-h] } = \lim \limits_{h \to 0} \frac{3}{2} = 2$

$\displaystyle \text{RHL of } \lim \limits_{x \to 3} \ \frac{x}{[x]}: \lim \limits_{x \to 3^+} \frac{x}{|x|}$

$\displaystyle \text{Let } x = 3+h, \text{ where } h \rightarrow 0$

$\displaystyle \Rightarrow \lim \limits_{h \to 0} \frac{3+h}{[3+h]} = \lim \limits_{h \to 0} \frac{3}{3} = 1$

$\displaystyle \text{Clearly, } \lim \limits_{x \to 3^-} \frac{x}{[x]} \neq \lim \limits_{x \to 3^+} \frac{x}{|x|} .$

$\displaystyle \text{Hence, } \lim \limits_{x \to 3} \frac{x}{|x|} \text{ does not exist.}$

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$\displaystyle \text{Question 19: Find } \lim \limits_{x \to \frac{5}{2}} \ [x] .$