Class 11 RD Sharma Solutions - Chapter 29 Limits - Exercise 29.1

In this section, we dive into Chapter 29 of the Class 11 RD Sharma textbook, which focuses on Limits. Exercise 29.1 is designed to introduce students to the fundamental concepts of limits, a crucial topic in calculus, helping them understand how to evaluate limits for different types of functions.

Class 11 RD Sharma Solutions - Chapter 29 Limits - Exercise 29.1

This section provides detailed solutions for Exercise 29.1 from Chapter 29 of the Class 11 RD Sharma textbook. These solutions are aimed at helping students grasp the basic principles of limits, laying the groundwork for more advanced topics in calculus.

Question 1. Show that Lim_x→0(x/|x|) does not exist.

Solution:

We have, Lim_x→0(x/|x|)

Now first we find left-hand limit:

= \lim_{x\to0^-}\frac{x}{|x|}

Let x = 0 - h, where h = 0

= \lim_{h\to0^-}(\frac{(0 - h)}{|0 - h|})

= \lim_{h\to0^-}(\frac{(- h)}{h})  

= -1

Now we find right-hand limit:

= \lim_{x\to0^+}\frac{x}{|x|}

So, let x = 0 + h, where h = 0

= \lim_{h\to0^+}(\frac{(0 + h)}{|0 + h|})

= \lim_{h\to0^+}(\frac{h}{h})

= 1

Left-hand limit ≠ Right-hand limit

So, Lim_x→0(x/|x|) does not exist.

Question 2. Find k so that Lim_x→0f(x), where f(x)= \begin{cases} 2x+3, \hspace{0.2cm}x≤2\\ x+k,\hspace{0.2cm}x>2 \end{cases}

Solution:

We have, f(x)= \begin{cases} 2x+3, \hspace{0.2cm}x≤2\\ x+k,\hspace{0.2cm}x>2 \end{cases}

Now first we find left-hand limit:

= \lim_{x\to0^-}(2x+3)

Let x = 2 - h, where h= 0.

=\lim_{h\to0^-}(2(2-h)+3)

= [2(2 - 0) + 3]

= 7

Now we find right-hand limit:

=\lim_{x\to0^+}f(x)

= \lim_{x\to0^+}(x+k)

Let x = 2 + h, where h = 0

= \lim_{h\to0^+}((2+h)+k)

= (2 + 0) + k

= (2 + k)

Here, Left-hand limit = Right-hand limit, so limit exists

So, (2 + k) = 7

k = 5

Question 3. Show that Lim_x→0(1/x) does not exist.

Solution:

We have to show that Lim_x→0(1/x) does not exists 

So for that 

First we find left-hand limit:

= \lim_{x\to0^-}\frac{1}{x}

Let x = 0 - h, where h = 0.

= \lim_{h\to0^-}(\frac{1}{(0-h)})

= -\lim_{h\to0^-}(\frac{1}{h})

= -∞

Now we find right-hand limit:

=\lim_{x\to0^+}\frac{1}{x}

Let x = 0 + h, where h = 0.

= \lim_{h\to0^+}(\frac{1}{|(0+h)|})

= \lim_{h\to0^+}(\frac{1}{h})

= ∞

Here, Left-hand limit ≠ Right-hand limit, so, Lim_x→0(1/x) does not exist.

Question 4. Let f(x) be a function defined by f(x)= \begin{cases} \frac{3x}{(|x|+2x)}, \hspace{0.2cm}x ≠ 0\\ 0,\hspace{0.2cm}x=0 \end{cases} . Show that lim_x→0 f(x) does not exist.

Solution:

We have, f(x)= \begin{cases} \frac{3x}{(|x|+2x)}, \hspace{0.2cm}x ≠ 0\\ 0,\hspace{0.2cm}x=0 \end{cases}

According to the question we have to show that lim_x→0 f(x) does not exist.

So for that 

First we find left-hand limit:

=\lim_{x\to0^-}\frac{3x}{|x|+2x}

Let x = 0 - h, where h = 0

=\lim_{h\to0^-}\frac{3(0-h)}{|0-h|+2(0-h)}

=\lim_{h\to0^-}(\frac{-3h}{h - 2h})

=\lim_{h\to0^-}(\frac{-3}{-1})

= 3

Now we find right-hand limit:

=\lim_{x\to0^+}\frac{3x}{|x|+2x}

Let x = 0 + h, where h = 0.

=\lim_{h\to0^+}\frac{3(0+h)}{|0+h|+2(0+h)}

=\lim_{h\to0^+}(\frac{3h}{3h})

=\lim_{h\to0^+}(\frac{3}{3})

= 1

Here, Left-hand limit ≠ Right-hand limit, so, lim_x→0 f(x) does not exist.

Question 5. Let f(x)= \begin{cases} x+1, \hspace{0.2cm}if \ x >0\\ x-1,\hspace{0.2cm}if \ x<0 \end{cases} , Prove that lim_x→0f(x) does not exist.

Solution:

We have, f(x)= \begin{cases} x+1, \hspace{0.2cm}if \ x >0\\ x-1,\hspace{0.2cm}if \ x<0 \end{cases}

And we have to prove that lim_x→0f(x) does not exist.

So for that 

First we find left-hand limit:

=\lim_{x\to0^-}(x-1)

Let x = 0 - h, where h = 0.

=\lim_{h\to0^-}((0-h)-1)

=\lim_{h\to0^-}(0-h-1)

= -1

Now we find right-hand limit:

=\lim_{x\to0^+}(x+1)

Let x = 0 + h, where h = 0.

=\lim_{h\to0^+}((0+h)+1)

=\lim_{h\to0^+}(0+h+1)

= 1

Here, Left-hand limit ≠ Right-hand limit, so, lim_x→0f(x) does not exist.

Question 6. Let f(x)= \begin{cases} x+5, \hspace{0.2cm}if \ x >0\\ x-4,\hspace{0.2cm}if \ x<0 \end{cases} , Prove that lim_x→0f(x) does not exist.

Solution:

We have,f(x)= \begin{cases} x+5, \hspace{0.2cm}if \ x >0\\ x-4,\hspace{0.2cm}if \ x<0 \end{cases}

And we have to prove that lim_x→0f(x) does not exist.

So for that 

First we find left-hand limit:

=\lim_{x\to0^-}(x-4)

Let x = 0 - h, where h = 0.

=\lim_{h\to0^-}((0-h)-4)

=\lim_{h\to0^-}(0-h-4)

= -4

Now we find right-hand limit:

=\lim_{x\to0^+}(x+5)

Let x = 0 + h, where h = 0.

=\lim_{h\to0^+}((0+h)+5)

=\lim_{h\to0^+}(0+h+5)

= 5

Here, Left-hand limit ≠ Right-hand limit, so, lim_x→0f(x) does not exist.

Question 7. Find lim_x→3f(x), where f(x)= \begin{cases} 4, \hspace{0.2cm}if \ x >3\\ x+1,\hspace{0.2cm}if \ x<3 \end{cases}

Solution:

We have, f(x)= \begin{cases} 4, \hspace{0.2cm}if \ x >3\\ x+1,\hspace{0.2cm}if \ x<3 \end{cases}

And we have to find lim_x→3f(x)

So for that 

First we find left-hand limit:

=\lim_{x\to3^-}(x+1)

Let x = 3 - h, where h = 0.

= \lim_{h\to0^-}((3-h)+1)  

= \lim_{h\to0^-}(4-h)  

= 4

Now we find right-hand limit:

=\lim_{x\to3^+}4

Let x = 3 + h, where h = 0.

=\lim_{h\to0^+}4

= 4

Here, Left-hand limit = Right-hand limit, 

Hence, lim_x→3f(x) = 4

Question 8(i). If f(x)= \begin{cases} 2x+3, \hspace{0.2cm}if \ x≤0\\ 3(x+1),\hspace{0.2cm}if \ x>0 \end{cases} , Find lim_x→0f(x).

Solution:

We have, f(x)= \begin{cases} 2x+3, \hspace{0.2cm}if \ x≤0\\ 3(x+1),\hspace{0.2cm}if \ x>0 \end{cases}

And we have to find lim_x→0f(x)

So for that 

First we find left-hand limit:

=\lim_{x\to0^-}(2x+3)

Let x = 0 - h, where h = 0.

= \lim_{h\to0^-}(2(0-h)+3)

= \lim_{h\to0^-}(3-2h)

= 3

Now we find right-hand limit:

=\lim_{x\to0^+}3(x+1)

Let x = 0 + h, where h = 0.

=\lim_{h\to0^+}3((0+h)+1)

=\lim_{h\to0^+}(3+3h)

= 3

Here, Left-hand limit = Right-hand limit, 

Hence, lim_x→0f(x) = 3

Question 8(ii). If f(x)= \begin{cases} 2x+3, \hspace{0.2cm}if \ x≤0\\ 3(x+1),\hspace{0.2cm}if \ x>0 \end{cases} , Find lim_x→1f(x).

Solution:

We have, f(x)= \begin{cases} 2x+3, \hspace{0.2cm}if \ x≤0\\ 3(x+1),\hspace{0.2cm}if \ x>0 \end{cases}

And we have to find lim_x→1f(x)

So for that 

First we find left-hand limit:

=\lim_{x\to1^-}(2x+3)

Let x = 1 - h, where h = 0.

= \lim_{h\to0^-}(2(1-h)+3)

= \lim_{h\to0^-}(5-2h)

= 5

Now we find right-hand limit:

=\lim_{x\to1^+}3(x+1)

Let x = 1 + h, where h = 0.

=\lim_{h\to0^+}(3((1+h)+1))

= \lim_{h\to0^+}(6+3h)

= 6

Here, Left-hand limit ≠ Right-hand limit, so lim_x→1f(x) does not exist.

Question 9. Find lim_x→1f(x) Where f(x)= \begin{cases} x^2-1, \hspace{0.2cm}if \ x≤1\\ -x^2-1,\hspace{0.2cm}if \ x>1 \end{cases}  

Solution:

We have, f(x)= \begin{cases} x^2-1, \hspace{0.2cm}if \ x≤1\\ -x^2-1,\hspace{0.2cm}if \ x>1 \end{cases}

And we have to find lim_x→1f(x)

So for that 

First we find left-hand limit:

=\lim_{x\to1^-}(x^2-1)

Let x = 1 - h, where h = 0.

= \lim_{h\to0^-}((1-h)^2-1)

=\lim_{h\to0^-}(1-2h-h^2-1)

= 0

Now we find right-hand limit:

=\lim_{x\to1^+}(-x^2-1)

Let x = 1 + h, where h = 0.

= \lim_{h\to0^+}(-(1+h)^2-1)

= \lim_{h\to0^+}(-1-2h-h^2-1)

= -2

Here, Left-hand limit ≠ Right-hand limit, so, lim_x→1f(x) does not exist.

Question 10. Evaluate lim_x→0f(x), where f(x)= \begin{cases} \frac{|x|}{x}, \hspace{0.2cm}if \ x≠1\\ 0,\hspace{0.2cm}if \ x=1 \end{cases}

Solution:

We have, f(x)= \begin{cases} \frac{|x|}{x}, \hspace{0.2cm}if \ x≠1\\ 0,\hspace{0.2cm}if \ x=1 \end{cases}

And we have to find lim_x→0f(x)

So for that 

First we find left-hand limit:

=\lim_{x\to0^-}\frac{|x|}{x}

Let x = 0 - h, where h = 0.

= \lim_{h\to0^-}\frac{|0-h|}{0-h}

= \lim_{h\to0^-}(\frac{h}{-h})

= -1

Now we find right-hand limit:

=\lim_{x\to0^+}\frac{|x|}{x}

Let x = 0 + h, where h = 0.

= \lim_{h\to0^+}\frac{|0+h|}{0+h}

= \lim_{h\to0^+}(\frac{h}{h})

= 1

Here, Left-hand limit ≠ Right-hand limit, so, lim_x→0f(x) does not exist.

Question 11. Let a₁, a₂,..........a_n be fixed real number such that f(x) = (x - a₁)(x - a₂)........(x-a_n). What is lim_x→a1f(x)? Compute lim_x→af(x).

Solution:

We have, f(x) = (x - a₁)(x - a₂)........(x - a_n)

\lim_{x→a_1}f(x)=\lim_{x→a_1}[(x-a_1)(x-a_2)........(x-a_n)]

Now, put x = a₁

= (a₁ - a₁)(a₁ - a₂)........(a₁ - a_n)

= 0

Now, lim_x→af(x) = lim_x→a[(x - a₁)(x - a₂)........(x - a_n)]

Now, put x = a

= (a - a₁)(a - a₂)........(a - a_n)

Hence, lim_x→af(x) = (a - a₁)(a - a₂)........(a - a_n)

Question 12. Find lim_x→1+[1/(x - 1)].

Solution:

We have to find lim_x→1+[1/(x - 1)]

=\lim_{x\to1^+}\frac{1}{x-1}

Let x = 1 + h, where h = 0.

=\lim_{h\to0^+}\frac{1}{(1+h)-1}

=\lim_{h\to0^+}\frac{1}{(h)}

= ∞

Hence, lim_x→1+[1/(x - 1)] = ∞

Question 13(i). Evaluate the following one-sided limits: lim_x→2+[(x - 3)/(x² - 4)]

Solution:

We have, \lim_{x\to2^+}\frac{x-3}{x^2-4}

Let x = 2 + h, where h = 0.

=\lim_{h\to0^+}(\frac{(2+h)-3}{(2+h)^2-4})  

= \lim_{h\to0^+}(\frac{h-1}{4+4h+h^2-4})

= \lim_{h\to0^+}(\frac{h-1}{4h+h^2})

= -∞

Question 13(ii). Evaluate the following one-sided limits: lim_x→2-[(x - 3)/(x² - 4)]

Solution:

We have,\lim_{x\to2^-}\frac{x-3}{x^2-4}

Let x = 2 - h, where h = 0.

= \lim_{h\to0^-}\frac{(2-h)-3}{(2-h)^2-4}

= \lim_{h\to0^-}\frac{(-h-1)}{4-4h+h^2-4}

= \lim_{h\to0^-}\frac{(-h-1)}{-4h+h^2}

= ∞

Question 13(iii). Evaluate the following one-sided limits: lim_x→0+[1/3x]

Solution:

We have, lim_x→0+[1/3x]

Let x = 0 + h, where h = 0.

= Lim_h→0+[1/3(0+h)]

= Lim_h→0+[1/(3h)]

= ∞

Question 13(iv). Evaluate the following one-sided limits: lim_x→-8+[2x/(x + 8)]

Solution:

We have, lim_x→-8+[2x/(x + 8)]

Let x = -8 + h, where h = 0.

= lim_x→0+[2(-8 + h)/(-8 + h + 8)]

= Lim_h→0+[(2h - 16)/(h)]

= -∞

Question 13(v). Evaluate the following one-sided limits: lim_x→0+[2/x^1/5]

Solution:

We have, lim_x→0+[2/x^1/5]

Let x = 0 + h, where h = 0.

= Lim_h→0+[2/(0 + h)^1/5]

= ∞

Question 13(vi). Evaluate the following one-sided limits: lim_x→(π/2)-[tanx]

Solution:

We have, lim_x→(π/2)-[tanx]

Let x = 0 - h, where h = 0.

= lim_h→0-[tan(π/2 - h)]

= lim_x→0-[cot h]

= ∞

Question 13(vii). Evaluate the following one-sided limits: lim_x→(-π/2)+[secx]

Solution:

We have, lim_x→(-π/2)+[secx]

Let x = 0 + h, where h = 0.

= lim_h→0+[secx(-π/2 + h)]

= lim_h→0+[cosec h]

= ∞

Question 13(viii). Evaluate the following one-sided limits: lim_x→0-[(x² - 3x + 2)/x³ - 2x²]

Solution:

We have, lim_x→0-[x² - 3x + 2/x³ - 2x²]

= Lim_x→0-[(x - 1)(x - 2)/x²(x - 2)]

= Lim_x→0-[(x - 1)/x²]

Let x = 0 - h, where h = 0.

= Lim_h→0-[(0 - h - 1)/(0 - h)²]

= -∞

Question 13(ix). Evaluate the following one-sided limits: lim_x→-2+[(x² - 1)/(2x + 4)]

Solution:

We have, lim_x→-2+[(x² - 1)/(2x + 4)]

Let x = -2 + h, where h = 0.

= Lim_h→-0⁺[(-2 + h)² - 1)/2(-2 + h) + 4]

= Lim_h→-0⁺[(-2 + h)² - 1)/(-4 + 4 + h)]

= (4 - 1)/0

= ∞

Question 13(x). Evaluate the following one-sided limits: lim_x→0-[2 - cotx]

Solution:

We have, lim_x→0-[2 - cotx]

Let x = 0 - h, where h = 0.

= Lim_h→0-[2 - cot(0 - h)]

= Lim_h→0-[2 + cot(h)]

= 2 + ∞

= ∞

Question 13(xi). Evaluate the following one-sided limits. lim_x→0-[1 + cosecx]

Solution:

We have, lim_x→0-[1 + cosecx]

Let x = 0 - h, where h = 0.

= Lim_h→0-[1 + cosec(0 - h)]

= Lim_h→0-[1 - cosec(h)]

= 1 - ∞

= -∞

Question 14. Show that Lim_x→0e^-1/x does not exist.

Solution:

Let, f(x) = Lim_x→0e^-1/x

So for that 

First we find left-hand limit:

= \lim_{x\to0^-}e^{-\frac{1}{x}}

Let x = 0 - h, where h = 0.

= \lim_{h\to0^-}e^{-\frac{1}{0-h}}

=\lim_{h\to0^-}e^{\frac{1}{h}}

= e^∞

= ∞

Now we find right-hand limit:

=\lim_{x\to0^+}e^{-\frac{1}{x}}

Let x = 0 + h, where h = 0.

= \lim_{h\to0^+}e^{-\frac{1}{0 + h}}

=\lim_{h\to0^+}e^{-\frac{1}{h}}

= e^-∞

= 0

Here, Left-hand limit ≠ Right-hand limit, so, Lim_x→0e^-1/x does not exist.

Question 15(i). Find Lim_x→2[x]  

Solution:

We have, Lim_x→2[x], where [] is Greatest Integer Function

So for that 

First we find left-hand limit:

=\lim_{x\to0^-}[x]

Let x = 2 - h, where h = 0.

= \lim_{h\to0^-}[2 - h]

= 1

Now we find right-hand limit:

= \lim_{x\to0^+}[x]

Let x = 2 + h, where h = 0.

= \lim_{h\to0^+}[2+h]

= 2

Here, Left-hand limit ≠ Right-hand limit, so, Lim_x→2[x] does not exist.

Question 15(ii). Find Lim_x→5/2[x]  

Solution:

We have, Lim_x→2[x], where [] is Greatest Integer Function

So for that 

First we find left-hand limit:

=\lim_{x\to\frac{5}{2}^-}[x]

Let x = 5/2 - h, where h = 0.

= \lim_{h\to0^-}[\frac{5}{2} - h]

= 2

Now we find right-hand limit:

=\lim_{x\to\frac{5}{2}^+}[x]

Let x = 5/2 + h, where h = 0.

= \lim_{h\to0^+}[\frac{5}{2}+h]

= 2

Here, Left-hand limit = Right-hand limit, so, Lim_x→5/2[x] = 2

Question 15(iii). Find Lim_x→1[x]  

Solution:

We have, Lim_x→1[x], where [] is Greatest Integer Function

So for that 

First we find left-hand limit:

=\lim_{x\to1^-}[x]

Let x = 1 - h, where h = 0.

= \lim_{h\to0^-}[1-h]

= 0

Now we find right-hand limit:

= \lim_{x\to1^+}[x]

Let x = 1 + h, where h = 0.

= \lim_{h\to0^+}[1+h]  

= 1

Here, Left-hand limit = Right-hand limit, so, Lim_x→1[x] does not exist.

Question 16. Prove that Lim_x→a+[x] = [a]. Also prove that Lim_x→1-[x] = 0.

Solution:

We have, \lim_{x\to a^+}[x]

Let x = a + h, where h = 0.

= Lim_h→0-[(a + h)]

= a

Also, \lim_{x\to1^-}[x]

Let x = 1 - h, where h = 0.

= Lim_h→0[(1 - h)]

= 0

Question 17. Show that Lim_x→2+(x/[x]) ≠ Lim_x→2-(x/[x]).

Solution:

We have to show Lim_x→2+(x/[x]) ≠ Lim_x→2-(x/[x])

So, R.H.L

We have, \lim_{x\to2^-}\frac{x}{[x]}  , where [] is greatest Integer Function

Let x = 2 - h, where h = 0.

= Lim_h→0-[(2 - h)/|[2 - h]]

= 2/1

= 2

Now, L.H.L 

We have, \lim_{x\to2^+}\frac{x}{[x]} , where [] is greatest Integer Function

Let x = 2 + h, where h = 0.

= Lim_h→0+[(2 + h)/|[2 + h]]

= 2/2

= 1

Hence, Left-hand limit≠Right-hand limit

Question 18. Find Lim_x→3+(x/[x]). Is it equal to Lim_x→3-(x/[x]) 

Solution:

We have, \lim_{x\to3^-}\frac{x}{[x]}   Where [] is Greatest Integer Function

Let x = 3 - h, where h = 0.

= Lim_h→0-[(3 - h)/|[3 - h]]

= 3/2

Also, \lim_{x\to3^+}\frac{x}{[x]}

Let x = 3 + h, where h = 0.

= Lim_h→0+[(3 + h)/|[3 + h]]

= 3/3

= 1

Hence, Left-hand limit≠Right-hand limit

Question 19. Find Lim_x→5/2[x]  

Solution:

We have to find Lim_x→5/2[x], where [] is Greatest Integer Function

So for that 

First we find left-hand limit:

= \lim_{x\to\frac{5}{2}^-}[x]

Let x = 5/2 - h, where h = 0.

= Lim_h→0-[(5/2 - h)]

= 2

Now we find right-hand limit:

=\lim_{x\to\frac{5}{2}^+}[x]

Let x = 5/2 + h, where h = 0.

= Lim_h→0+[(5/2+h)]

= 2

Hence, Left-hand limit = Right-hand limit, so Lim_x→5/2[x]  = 2

Question 20. Evaluate Lim_x→2f(x), where f(x)= \begin{cases} x-[x], \hspace{0.2cm}x<2\\  4 , \hspace{0.2cm}x=2\\ x+k,\hspace{0.2cm}x>2 \end{cases}

Solution:

We have, f(x)= \begin{cases} x-[x], \hspace{0.2cm}x<2\\  4 , \hspace{0.2cm}x=2\\ x+k,\hspace{0.2cm}x>2 \end{cases}

We have to find Lim_x→2f(x)

So for that 

First we find left-hand limit:

= \lim_{x\to0^-}(x-[x])

Let x = 2 - h, where h = 0.

= Lim_h→0-{(2 - h) - [2 - h]}

= 2 - 1

= 1

Now we find right-hand limit:

=\lim_{x\to2^+}(3x-5)

Let x = 2 + h, where h = 0.

= Lim_h→0-[3(2 + h) - 5]

= 6 - 5

= 1

Hence, Left-hand limit = Right-hand limit, so, Lim_x→2f(x) = 1

Question 21. Show that Lim_x→0sin(1/x) does not exist.

Solution:

Let, f(x) = Lim_x→0sin(1/x)

First we find left-hand limit:

= \lim_{x\to0^-}sin\frac{1}{x}

Let x = 0 - h, where h = 0.

= Lim_h→0sin[1/(0 - h)]

= -Lim_h→0sin[1/(h)]

An oscillating number lies between -1 to +1.

So left hand limit does not exists.

Similarly, right-hand limit is also oscillating.

So, Lim_x→0sin(1/x) does not exist.

Question 22. Let f(x)= \begin{cases} \frac{kcosx}{\pi-2x}, \hspace{0.2cm}where \ x \neq \frac{\pi}{2}\\ 3,\hspace{0.2cm}where \ x =\frac{\pi}{2} \end{cases}  and if lim _x→π​/2 f(x) = f(π/2), find the value of k.

Solution:

We have f(x)= \begin{cases} \frac{kcosx}{\pi-2x}, \hspace{0.2cm}where \ x \neq \frac{\pi}{2}\\ 3,\hspace{0.2cm}where \ x =\frac{\pi}{2} \end{cases} 

First we find left-hand limit:

= \lim_{x\to\frac{\pi}{2}^-}\frac{kcosx}{\pi-2x}

Let x = π/2 - h, where h = 0.

=\lim_{h\to0^-}\frac{kcos(\frac{\pi}{2}-h)}{\pi-2(\frac{\pi}{2}-h)}

= k cos(π/2 - π/2)/π

= k/π

Now we find right-hand limit:

\lim_{x\to\frac{\pi}{2}^+}\frac{kcosx}{\pi-2x}

Let x = π/2 + h, where h = 0.

=\lim_{h\to0^+}\frac{kcos(\frac{\pi}{2}+h)}{\pi-2(\frac{\pi}{2}+h)}

= k cos(π/2 + π/2)/-π

= k/π

Hence, Left-hand limit = Right-hand limit, so

lim _x→π​/2 f(x) = f(π/2)

k/π = 3 

k = 3π

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