NCERT Solutions Class 11 – Chapter 12 Limits And Derivatives – Exercise 12.1

Evaluate the following limits in Exercises 1 to 22.

Question 1: [Tex]\lim_{x \to 3} x+3[/Tex]

Solution:

In [Tex]\lim_{x \to 3} x+3[/Tex], as x⇢3 

Put x = 3, we get

[Tex]\lim_{x \to 3} x+3[/Tex] = 3+3 

= 6

Question 2: [Tex]\lim_{x \to \pi} (x-\frac{22}{7})[/Tex]

Solution:

In [Tex]\lim_{x \to \pi} (x-\frac{22}{7})[/Tex], as x⇢π

Put x = π, we get

[Tex]\lim_{x \to \pi} (x-\frac{22}{7}) = (π-\frac{22}{7}) [/Tex]

= [Tex](π-\frac{22}{7})[/Tex]

Question 3: [Tex]\lim_{r \to 1} \pi r^2[/Tex]

Solution:

In [Tex]\lim_{r \to 1} \pi r^2[/Tex], as r⇢1

Put r = 1, we get

[Tex]\lim_{r \to 1} \pi r^2 = \pi (1)^2 [/Tex]

= π

Question 4: [Tex]\lim_{x \to 4} (\frac{4x+3}{x-2})[/Tex]

Solution:

In [Tex]\lim_{x \to 4} (\frac{4x+3}{x-2})[/Tex], as x⇢4

Put x = 4, we get

[Tex]\lim_{x \to 4} (\frac{4x+3}{x-2}) = \frac{4(4)+3}{4-2} [/Tex]

= [Tex]\frac{19}{2}[/Tex]

Question 5: [Tex]\lim_{x \to -1} (\frac{x^{10}+x^5+1}{x-1})[/Tex]

Solution:

In [Tex]\lim_{x \to -1} (\frac{x^{10}+x^5+1}{x-1})[/Tex], as x⇢-1

Put x = -1, we get

[Tex]\lim_{x \to -1} (\frac{x^{10}+x^5+1}{x-1}) = \frac{(-1)^{10}+(-1)^5+1}{-1-1}[/Tex]

= [Tex]\frac{1-1+1}{-2}[/Tex]

= [Tex]\frac{-1}{2}[/Tex]

Question 6: [Tex]\lim_{x \to 0} \frac{(x+1)^5-1}{x}[/Tex]

Solution:

In [Tex]\lim_{x \to 0} \frac{(x+1)^5-1}{x}[/Tex], as x⇢0

Put x = 0, we get

[Tex]\lim_{x \to 0} \frac{(x+1)^5-1}{x} = \frac{(0+1)^5-1}{0} = \frac{0}{0}[/Tex]

As, this limit becomes undefined

Now, let’s take x+1=p and x = p-1, to make it equivalent to theorem.

[Tex]\mathbf{\lim_{x \to a} \frac{x^n-a^n}{x-a} = na^{n-1}}[/Tex]

As, x⇢0 ⇒ p⇢1

[Tex]\lim_{p \to 1} \frac{(p)^5-1}{p-1} = \lim_{p \to 1} \frac{(p)^5-1^5}{p-1}[/Tex]

Here, n=5 and a = 1.

[Tex]\lim_{p \to 1} \frac{(p)^5-1}{p-1} = 5(1)^{5-1} [/Tex]

= 5(1)⁴ 

= 5

Question 7: [Tex]\lim_{x \to 2} \frac{3x^2-x-10}{x^2-4}[/Tex]

Solution:

In [Tex]\lim_{x \to 2} \frac{3x^2-x-10}{x^2-4}[/Tex], as x⇢2

Put x = 2, we get

[Tex]\lim_{x \to 2} \frac{3x^2-x-10}{x^2-4} = \frac{3(2)^2-2-10}{2^2-4} = \frac{0}{0}[/Tex]

As, this limit becomes undefined

Now, let’s Factorise the numerator and denominator, we get

[Tex]\lim_{x \to 2} \frac{3x^2-6x+5x-10}{x^2-4}[/Tex]

= [Tex]\lim_{x \to 2} \frac{(3x+5)(x-2)}{(x+2)(x-2)}[/Tex]

Cancelling (x-2), we have

= [Tex]\lim_{x \to 2} \frac{(3x+5)}{(x+2)}[/Tex]

Put x = 2, we get

[Tex]\lim_{x \to 2} \frac{(3x+5)}{(x+2)} = \frac{(3(2)+5)}{(2+2)} [/Tex]

= [Tex]\frac{11}{4} [/Tex]

Question 8: [Tex]\lim_{x \to 3} \frac{x^4-81}{2x^2-5x-3}[/Tex]

Solution:

In [Tex]\lim_{x \to 3} \frac{x^4-81}{2x^2-5x-3}[/Tex], as x⇢3

Put x = 3, we get

[Tex]\lim_{x \to 3} \frac{x^4-81}{2x^2-5x-3} = \frac{(3)^4-81}{2(3)^2-5(3)-3} = \frac{0}{0}[/Tex]

As, this limit becomes undefined

Now, let’s Factorise the numerator and denominator, we get

[Tex]\lim_{x \to 3} \frac{(x^2)^2-9^2}{2x^2-6x+x-3}[/Tex]

= [Tex]\lim_{x \to 3} \frac{(x^2-9)(x^2+9)}{(2x+1)(x-3)}[/Tex]

= [Tex]\lim_{x \to 3} \frac{(x+3)(x-3)(x^2+9)}{(2x+1)(x-3)}[/Tex]

Cancelling (x-3), we have

= [Tex]\lim_{x \to 3} \frac{(x+3)(x^2+9)}{(2x+1)}[/Tex]

Put x = 3, we get

= [Tex]\lim_{x \to 3} \frac{(x+3)(x^2+9)}{(2x+1)} = \frac{(3+3)(3^2+9)}{(2(3)+1)}[/Tex]

= [Tex]\frac{9\times 18}{7} [/Tex]

= [Tex]\frac{108}{7} [/Tex]

Question 9: [Tex]\lim_{x \to 0} \frac{ax+b}{cx+1}[/Tex]

Solution:

In [Tex]\lim_{x \to 0} \frac{ax+b}{cx+1}[/Tex], as x⇢0

Put x = 0, we get

[Tex]\lim_{x \to 0} \frac{ax+b}{cx+1} = \frac{a(0)+b}{c(0)+1}[/Tex]

= b

Question 10: [Tex]\lim_{z \to 1} \frac{z^{\frac{1}{3}}-1}{z^{\frac{1}{6}}-1}[/Tex]

Solution:

In [Tex]\lim_{z \to 1} \frac{z^{\frac{1}{3}}-1}{z^{\frac{1}{6}}-1}[/Tex], as z⇢1

Put z = 1, we get

[Tex]\lim_{z \to 1} \frac{z^{\frac{1}{3}}-1}{z^{\frac{1}{6}}-1} = \frac{1^{\frac{1}{3}}-1}{1^{\frac{1}{6}}-1} = \frac{0}{0}[/Tex]

Let’s take [Tex]z^{\frac{1}{6}}  [/Tex] = p and [Tex]z^{\frac{1}{3}}  [/Tex] = p²,

As, z⇢1 ⇒ p⇢1

= [Tex]\lim_{p \to 1} \frac{p^2-1}{p-1}[/Tex]

Now, let’s Factorise the numerator, we get

= [Tex]\lim_{p \to 1} \frac{(p-1)(p+1)}{p-1}[/Tex]

Cancelling (p-1), we have

= [Tex]\lim_{p \to 1} (p+1)[/Tex]

Put p = 1, we get

[Tex]\lim_{p \to 1} (p+1) = 1+1[/Tex]

= 2

Question 11: [Tex]\lim_{x \to 1} \frac{ax^2+bx+c}{cx^2+bx+a},\hspace{0.1cm}a+b+c\neq0[/Tex]

Solution:

In [Tex]\lim_{x \to 1} \frac{ax^2+bx+c}{cx^2+bx+a}[/Tex], as x⇢1

Put x = 1, we get

[Tex]\lim_{x \to 1} \frac{ax^2+bx+c}{cx^2+bx+a} = \frac{a(1)^2+b(1)+c}{c(1)^2+b(1)+a}[/Tex]

= [Tex]\frac{a+b+c}{c+b+a}[/Tex]

= 1 (As it is given a+b+c≠0)

Question 12: [Tex]\lim_{x \to -2} \frac{\frac{1}{x} + \frac{1}{2}}{x+2}[/Tex]

Solution:

In [Tex]\lim_{x \to -2} \frac{\frac{1}{x} + \frac{1}{2}}{x+2}[/Tex], as x⇢-2

Firstly, lets simplify the equation

[Tex]\frac{\frac{1}{x} + \frac{1}{2}}{x+2} = \frac{\frac{2+x}{2x}}{x+2}[/Tex]

Cancelling (x+2),we get

[Tex]\lim_{x \to -2} \frac{1}{2x}[/Tex]

Put x = -2, we get

[Tex]\lim_{x \to -2} \frac{1}{2x} = \frac{1}{2(-2)}[/Tex]

= [Tex]\frac{-1}{4}[/Tex]

Question 13: [Tex]\lim_{x \to 0} \frac{sin\hspace{0.1cm}ax}{bx}[/Tex]

Solution:

In [Tex]\lim_{x \to 0} \frac{sin\hspace{0.1cm}ax}{bx}[/Tex], as x⇢0

Put x = 0, we get

[Tex]\lim_{x \to 0} \frac{sin\hspace{0.1cm}ax}{bx} = \frac{sin\hspace{0.1cm}a(0)}{b(0)} = \frac{0}{0}[/Tex]

As, this limit becomes undefined

Now, let’s multiply and divide the equation by a, to make it equivalent to theorem.

[Tex]\mathbf{\lim_{x \to 0} \frac{sin \hspace{0.1cm}x}{x} = 1}[/Tex]

Hence, we have

[Tex]\lim_{x \to 0} \frac{sin\hspace{0.1cm}ax}{bx} \times \frac{a}{a}[/Tex]

 = [Tex]\lim_{x \to 0} \frac{sin\hspace{0.1cm}ax}{ax} \times \frac{a}{b}[/Tex]

= [Tex]\frac{a}{b} \lim_{x \to 0} \frac{sin\hspace{0.1cm}ax}{ax}[/Tex]

As x⇢0, then ax⇢0

= [Tex]\frac{a}{b} \lim_{ax \to 0} \frac{sin\hspace{0.1cm}ax}{ax}[/Tex]

By using the theorem, we get

 = [Tex]\frac{a}{b} . 1[/Tex]

= [Tex]\frac{a}{b}[/Tex]

Question 14: [Tex]\lim_{x \to 0} \frac{sin \hspace{0.1cm}ax}{sin \hspace{0.1cm}bx},\hspace{0.1cm}a,b\neq0[/Tex]

Solution:

In [Tex]\lim_{x \to 0} \frac{sin \hspace{0.1cm}ax}{sin \hspace{0.1cm}bx}[/Tex], as x⇢0

Put x = 0, we get

[Tex]\lim_{x \to 0} \frac{sin \hspace{0.1cm}ax}{sin \hspace{0.1cm}bx} = \frac{sin \hspace{0.1cm}a(0)}{sin \hspace{0.1cm}b(0)} = \frac{0}{0}[/Tex]

As, this limit becomes undefined

Now, let’s multiply and divide the numerator by ax and denominator by bx to make it equivalent to theorem.

[Tex]\mathbf{\lim_{x \to 0} \frac{sin \hspace{0.1cm}x}{x} = 1}[/Tex]

Hence, we have

[Tex]\lim_{x \to 0} \frac{\frac{sin \hspace{0.1cm}(ax) \times ax}{ax}}{\frac{sin \hspace{0.1cm}(bx) \times bx}{bx}}[/Tex]

= [Tex]\frac{a}{b}\lim_{x \to 0} \frac{\frac{sin \hspace{0.1cm}ax}{ax}}{\frac{sin \hspace{0.1cm}bx}{bx}}[/Tex]

= [Tex]\frac{a}{b} \frac{\lim_{x \to 0}\frac{sin \hspace{0.1cm}ax}{ax}}{\lim_{x \to 0}\frac{sin \hspace{0.1cm}bx}{bx}}[/Tex]

By using the theorem, we get

= [Tex]\frac{a}{b} . 1 .1[/Tex]

= [Tex]\frac{a}{b}[/Tex]

Question 15: [Tex]\lim_{x \to \pi} \frac{sin(\pi-x)}{\pi(\pi-x)}[/Tex]

Solution:

In [Tex]\lim_{x \to \pi} \frac{sin(\pi-x)}{\pi(\pi-x)}[/Tex], as x⇢π

Put x = π, we get

[Tex]\lim_{x \to \pi} \frac{sin(\pi-x)}{\pi(\pi-x)} = \frac{sin(\pi-\pi)}{\pi(\pi-\pi)} = \frac{0}{0}[/Tex]

As, this limit becomes undefined

Now, let’s take π-x=p

As, x⇢π ⇒ p⇢0

[Tex]\mathbf{\lim_{x \to 0} \frac{sin \hspace{0.1cm}x}{x} = 1}[/Tex]

= [Tex]\lim_{p \to 0} \frac{sin p}{p\pi}[/Tex]

= [Tex]\frac{1}{\pi} \lim_{p \to 0} \frac{sin\hspace{0.1cm} p}{p}[/Tex]

By using the theorem, we get

= [Tex]1. \frac{1}{\pi}[/Tex]

= [Tex]\frac{1}{\pi}[/Tex]

Question 16: [Tex]\lim_{x \to 0} \frac{cos\hspace{0.1cm}x}{(\pi-x)}[/Tex]

Solution:

In [Tex]\lim_{x \to 0} \frac{cos\hspace{0.1cm}x}{(\pi-x)}[/Tex], as x⇢0

Put x = 0, we get

[Tex]\lim_{x \to 0} \frac{cos\hspace{0.1cm}x}{(\pi-x)} = \frac{cos\hspace{0.1cm}0}{(\pi-0)}[/Tex]

= [Tex]\frac{1}{\pi}[/Tex]

Question 17:[Tex]\lim_{x \to 0} \frac{cos \hspace{0.1cm}2x-1}{cos \hspace{0.1cm}x-1}[/Tex]

Solution:

In[Tex]\lim_{x \to 0} \frac{cos \hspace{0.1cm}2x-1}{cos \hspace{0.1cm}x-1} [/Tex], as x⇢0

As we know, cos 2θ = 1-2sin²θ

Substituting the values, we get

[Tex]\lim_{x \to 0} \frac{1-2sin^2x-1}{1-2sin^2(\frac{x}{2})-1}[/Tex]

=[Tex]\lim_{x \to 0} \frac{sin^2x}{sin^2(\frac{x}{2})}[/Tex]

Put x = 0, we get

[Tex]\lim_{x \to 0} \frac{sin^2x}{sin^2(\frac{x}{2})} = \frac{0}{0}[/Tex]

As, this limit becomes undefined

Now, let’s multiply and divide the numerator by x² and denominator by[Tex](\frac{x}{2})^2 [/Tex]to make it equivalent to theorem.

[Tex]\mathbf{\lim_{x \to 0} \frac{sin x}{x} = 1}[/Tex]

Hence, we have

[Tex]\lim_{x \to 0} \frac{\frac{sin^2x \times x^2}{x^2}}{\frac{sin^2(\frac{x}{2}) \times (\frac{x}{2})^2}{(\frac{x}{2})^2}}[/Tex]

=[Tex]\lim_{x \to 0} \frac{(\frac{sin \hspace{0.1cm}x}{x})^2\times x^2}{(\frac{sin \hspace{0.1cm}(\frac{x}{2})}{\frac{x}{2}})^2\times (\frac{x}{2})^2}[/Tex]

=[Tex]\frac{(\lim_{x \to 0}\frac{sin \hspace{0.1cm}x}{x})^2\times\lim_{x \to 0} x^2}{(\lim_{x \to 0}\frac{sin \hspace{0.1cm}(\frac{x}{2})}{\frac{x}{2}})^2\times\lim_{x \to 0} (\frac{x}{2})^2}[/Tex]

By using the theorem, we get

=[Tex]\frac{(1)^2\times\lim_{x \to 0} x^2}{(1)^2\times\lim_{x \to 0} (\frac{x^2}{4})}[/Tex]

=[Tex]\lim_{x \to 0}\frac{x^2}{(\frac{x^2}{4})}[/Tex]

=[Tex]\lim_{x \to 0}(4)[/Tex]

= 4

Question 18:[Tex]\lim_{x \to 0} \frac{ax+xcos \hspace{0.1cm}x}{bsin \hspace{0.1cm}x}[/Tex]

Solution:

In[Tex]\lim_{x \to 0} \frac{ax+xcos \hspace{0.1cm}x}{bsin \hspace{0.1cm}x} [/Tex], as x⇢0

Put x = 0, we get

[Tex]\lim_{x \to 0} \frac{ax+xcos \hspace{0.1cm}x}{bsin \hspace{0.1cm}x} = \frac{0}{0}[/Tex]

As, this limit becomes undefined

Now, let’s simplify the equation to make it equivalent to theorem.

[Tex]\mathbf{\lim_{x \to 0} \frac{sin x}{x} = 1}[/Tex]

Hence, we have

[Tex]\lim_{x \to 0} \frac{x(a+cos \hspace{0.1cm}x)}{bsin \hspace{0.1cm}x}[/Tex]

=[Tex]\frac{1}{b}\times \lim_{x \to 0} \frac{x}{sin \hspace{0.1cm}x}\times \lim_{x \to 0} (a+cos \hspace{0.1cm}x)[/Tex]

By using the theorem, we get

=[Tex]\frac{1}{b}\times 1\times \lim_{x \to 0} (a+cos \hspace{0.1cm}x)[/Tex]

=[Tex]\frac{1}{b}\times a[/Tex]

Putting x=0, we have

=[Tex]\frac{a}{b}[/Tex]

Question 19:[Tex]\lim_{x \to 0} x sec\hspace{0.1cm}x[/Tex]

Solution:

In[Tex]\lim_{x \to 0} x sec\hspace{0.1cm}x [/Tex], as x⇢0

Put x = 0, we get

[Tex]\lim_{x \to 0} x sec\hspace{0.1cm}x [/Tex]= 0 ×1

= 0

Question 20:[Tex]\lim_{x \to 0} \frac{sin \hspace{0.1cm}ax+bx}{ax+sin \hspace{0.1cm}bx}[/Tex]

Solution:

In[Tex]\lim_{x \to 0} \frac{sin \hspace{0.1cm}ax+bx}{ax+sin \hspace{0.1cm}bx} [/Tex], as x⇢0

Put x = 0, we get

[Tex]\lim_{x \to 0} \frac{sin \hspace{0.1cm}ax+bx}{ax+sin \hspace{0.1cm}bx} = \frac{0}{0}[/Tex]

As, this limit becomes undefined

Now, let’s simplify the equation to make it equivalent to theorem.

[Tex]\mathbf{\lim_{x \to 0} \frac{sin x}{x} = 1}[/Tex]

Hence, we can write the equation as follows:

[Tex]\lim_{x \to 0} \frac{\frac{sin \hspace{0.1cm}(ax)\times ax}{ax}+bx}{ax+\frac{sin \hspace{0.1cm}(bx)\times bx}{bx}}[/Tex]

=[Tex]\frac{\lim_{x \to 0}\frac{sin \hspace{0.1cm}(ax)}{ax}\times \lim_{x \to 0}ax+\lim_{x \to 0}bx}{\lim_{x \to 0}ax+\lim_{x \to 0}\frac{sin \hspace{0.1cm}(bx)}{bx}\times\lim_{x \to 0} bx}[/Tex]

By using the theorem, we get

=[Tex]\frac{1\times \lim_{x \to 0}ax+\lim_{x \to 0}bx}{\lim_{x \to 0}ax+1\times\lim_{x \to 0} bx}[/Tex]

=[Tex]\frac{\lim_{x \to 0}ax+\lim_{x \to 0}bx}{\lim_{x \to 0}ax+\lim_{x \to 0} bx}[/Tex]

=[Tex]\lim_{x \to 0} \frac{ax+bx}{ax+bx}[/Tex]

=[Tex]\lim_{x \to 0} 1[/Tex]

Putting x=0, we have

= 1

Question 21:[Tex]\lim_{x \to 0} (cosec\hspace{0.1cm}x-cot\hspace{0.1cm}x)[/Tex]

Solution:

In[Tex]\lim_{x \to 0} (cosec\hspace{0.1cm}x-cot\hspace{0.1cm}x) [/Tex], as x⇢0

By simplification, we get

[Tex]\lim_{x \to 0} (\frac{1}{sin\hspace{0.1cm}x}-\frac{cos\hspace{0.1cm}x}{sin\hspace{0.1cm}x})[/Tex]

[Tex]\lim_{x \to 0} (\frac{1-cos\hspace{0.1cm}x}{sin\hspace{0.1cm}x})[/Tex]

Put x = 0, we get

[Tex]\lim_{x \to 0} (\frac{1-cos\hspace{0.1cm}x}{sin\hspace{0.1cm}x}) = \frac{0}{0}[/Tex]

As, this limit becomes undefined

Now, let’s simplify the equation to make it equivalent to theorem:

[Tex]\mathbf{\lim_{x \to 0} \frac{sin x}{x} = 1}[/Tex]

By using the trigonometric identities,

cos 2θ = 1-2sin²θ

sin 2θ = 2 sinθ cosθ

Hence, we can write the equation as follows:

[Tex]\lim_{x \to 0} (\frac{2sin^2(\frac{x}{2})}{2 sin(\frac{x}{2})cos(\frac{x}{2})})[/Tex]

=[Tex]\lim_{x \to 0} (\frac{sin(\frac{x}{2})}{cos(\frac{x}{2})})[/Tex]

=[Tex]\lim_{x \to 0} tan(\frac{x}{2})[/Tex]

Putting x=0, we have

= 0

Question 22:[Tex]\lim_{x \to \frac{\pi}{2}} \frac{tan \hspace{0.1cm}2x}{x-\frac{\pi}{2}}[/Tex]

Solution:

In[Tex]\lim_{x \to \frac{\pi}{2}} \frac{tan \hspace{0.1cm}2x}{x-\frac{\pi}{2}} [/Tex], as x⇢[Tex]\frac{\pi}{2}[/Tex]

Put x =[Tex]\frac{\pi}{2} [/Tex], we get

[Tex]\lim_{x \to \frac{\pi}{2}} \frac{tan \hspace{0.1cm}2x}{x-\frac{\pi}{2}} = \frac{0}{0}[/Tex]

As, this limit becomes undefined

Now, let’s simplify the equation :

Let’s take[Tex]x-\frac{\pi}{2}=p[/Tex]

As, x⇢[Tex]\frac{\pi}{2} [/Tex]⇒ p⇢0

Hence, we can write the equation as follows:

[Tex]\lim_{p \to 0} \frac{tan \hspace{0.1cm}2(p+\frac{\pi}{2})}{p}[/Tex]

=[Tex]\lim_{p \to 0} \frac{tan \hspace{0.1cm}(2p+\pi)}{p}[/Tex]

=[Tex]\lim_{p \to 0} \frac{tan \hspace{0.1cm}(2p)}{p} [/Tex](As tan (π+θ) = tan θ)

=[Tex]\lim_{p \to 0} \frac{\frac{sin \hspace{0.1cm}(2p)}{cos\hspace{0.1cm}(2p)}}{p}[/Tex]

=[Tex]\lim_{p \to 0} \frac{sin \hspace{0.1cm}(2p)}{cos\hspace{0.1cm}(2p) \times p}[/Tex]

Now, let’s multiply and divide the equation by 2 to make it equivalent to theorem

[Tex]\mathbf{\lim_{x \to 0} \frac{sin x}{x} = 1}[/Tex]

=[Tex]\lim_{p \to 0} \frac{sin \hspace{0.1cm}(2p)}{cos\hspace{0.1cm}(2p) \times p} \times \frac{2}{2}[/Tex]

=[Tex]\lim_{p \to 0} \frac{2 sin \hspace{0.1cm}(2p)}{cos\hspace{0.1cm}(2p) \times 2p}[/Tex]

As p⇢0, then 2p⇢0

=[Tex]2. \lim_{2p \to 0} \frac{sin \hspace{0.1cm}(2p)}{2p} \times \lim_{p \to 0}\frac{1}{cos\hspace{0.1cm}(2p)}[/Tex]

Using the theorem and putting p=0, we have

= 2×1×1

= 2

Question 23: Find[Tex]\lim_{x \to 0} f(x) [/Tex]and[Tex]\lim_{x \to 1} f(x) [/Tex], where[Tex]f(x)= \begin{cases} 2x+3, \hspace{0.2cm}x\leq0\\ 3(x+1),\hspace{0.2cm}x>0 \end{cases}[/Tex]

Solution:

Let’s calculate, the limits when x⇢0

Here,

Left limit =[Tex]\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (2x+3)\\ = 2(0)+3\\ =3[/Tex]

Right limit =[Tex]\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} 3(x+1)\\ = 3(0+1)\\ =3[/Tex]

Limit value =[Tex]\lim_{x \to 0} f(x) = \lim_{x \to 0} (2x+3)\\ = 2(0)+3\\ =3[/Tex]

Hence,[Tex]\lim_{x \to 0^-} f(x)= \lim_{x \to 0} f(x)=\lim_{x \to 0^+} f(x)=3 [/Tex], then limit exists

Now, let’s calculate, the limits when x⇢1

Here,

Left limit =[Tex]\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} 3(x+1)\\= 3(1+1)\\=6[/Tex]

Right limit =[Tex]\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} 3(x+1)\\= 3(1+1)\\=6[/Tex]

Limit value =[Tex]\lim_{x \to 1} f(x) = \lim_{x \to 1} 3(x+1)\\= 3(1+1)\\=6[/Tex]

Hence,[Tex]\lim_{x \to 1^-} f(x)= \lim_{x \to 1} f(x)=\lim_{x \to 1^+} f(x) = 6 [/Tex], then limit exists

Question 24: Find[Tex]\lim_{x \to 1} f(x)[/Tex], where[Tex]f(x)= \begin{cases} x^2-1, \hspace{0.2cm}x\leq1\\ -x^2-1,\hspace{0.2cm}x>1 \end{cases}[/Tex]

Solution:

Let’s calculate, the limits when x⇢1

Here,

Left limit =[Tex]\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x^2-1)\\= 1^2-1\\=0[/Tex]

Right limit =[Tex]\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (-x^2-1)\\= -1^2-1\\=-2[/Tex]

As,[Tex]\lim_{x \to 1^-} f(x) \neq \lim_{x \to 1^+} f(x)[/Tex]

Hence, limit does not exists when x⇢1.

Question 25: Evaluate[Tex]\lim_{x \to 0} f(x) [/Tex], where[Tex]f(x)= \begin{cases} \frac{|x|}{x}, \hspace{0.2cm}x\neq0\\ 0,\hspace{0.2cm}x=0 \end{cases}[/Tex]

Solution:

Let’s calculate, the limits when x⇢0

Here,

As, we know that mod function works differently.

In |x-0|, |x|=x when x>0 and |x|=-x when x<0

Left limit =[Tex]\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{|x|}{x}\\= \frac{-x}{x}\\=-1[/Tex]

Right limit =[Tex]\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{|x|}{x}\\= \frac{x}{x}\\=1[/Tex]

As,[Tex]\lim_{x \to 0^-} f(x) \neq \lim_{x \to 0^+} f(x)[/Tex]

Hence, limit does not exists when x⇢0.

Question 26: Find [Tex]\lim_{x \to 0} f(x)[/Tex], where[Tex]f(x)= \begin{cases} \frac{x}{|x|}, \hspace{0.2cm}x\neq0\\ 0,\hspace{0.2cm}x=0 \end{cases}[/Tex]

Solution:

Let’s calculate, the limits when x⇢0

Here,

As, we know that mod function works differently.

In |x-0|, |x|=x when x>0 and |x|=-x when x<0

Left limit =[Tex]\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{x}{|x|}\\= \frac{x}{-x}\\=-1[/Tex]

Right limit =[Tex]\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{x}{|x|}\\= \frac{x}{x}\\=1[/Tex]

As,[Tex]\lim_{x \to 0^-} f(x) \neq \lim_{x \to 0^+} f(x)[/Tex]

Hence, limit does not exists when x⇢0.

Question 27: Find[Tex]\lim_{x \to 5} f(x) [/Tex], where f(x)=|x|-5.

Solution:

Let’s calculate, the limits when x⇢5

Here,

As, we know that mod function works differently.

In |x-0|, |x|=x when x>0 and |x|=-x when x<0

Left limit =[Tex]\lim_{x \to 5^-} f(x) = \lim_{x \to 5^-} |x|-5\\= x-5\\=5-5\\=0[/Tex]

Right limit =[Tex]\lim_{x \to 5^+} f(x) = \lim_{x \to 5^+} |x|-5\\= x-5\\=5-5\\=0[/Tex]

Hence,[Tex]\lim_{x \to 5^-} f(x)= \lim_{x \to 5} f(x)=\lim_{x \to 5^+} f(x) = 0 [/Tex], then limit exists

Question 28: Suppose[Tex]f(x)= \begin{cases} a+bx, \hspace{0.2cm}x<1\\ 4,\hspace{0.2cm}x=1\\ b-ax,\hspace{0.2cm}x>1 \end{cases} [/Tex]and if[Tex]\lim_{x \to 1} f(x) = f(1) [/Tex]what are possible values of a and b?

Solution:

As, it is given[Tex]\lim_{x \to 1} f(x) = f(1)[/Tex]

Let’s calculate, the limits when x⇢1

Here,

Left limit =[Tex]\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} a+bx\\= a+b(1)\\=a+b[/Tex]

Right limit =[Tex]\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} b-ax\\= b-a(1)\\=b-a[/Tex]

Limit value f(1) = 4

So, as limit exists then it should satisfy

[Tex]\lim_{x \to 1^-} f(x)= \lim_{x \to 1} f(x)=\lim_{x \to 1^+} f(x) = f(1) = 4[/Tex]

Hence, a+b = 4 and b-a = 4

Solving these equation, we get

a = 0 and b = 4

Question 29: Let a₁, a₂, . . ., a_n be fixed real numbers and define a function

f(x) = (x-a₁) (x-a₂)………… (x-an).

What is[Tex]\lim_{x \to a_1} f(x) [/Tex]? For some a ≠ a₁, a₂, …, an, compute[Tex]\lim_{x \to a} f(x)[/Tex].

Solution:

Here, f(x) = (x-a₁) (x-a₂)………… (x-a_n).

Then,[Tex]\lim_{x \to a_1} f(x) = \lim_{x \to a_1} (x-a_1) (x-a_2)………… (x-a_n)[/Tex]

=[Tex]\lim_{x \to a_1} (x-a1) \lim_{x \to a_1}(x-a_2)………… \lim_{x \to a_1}(x-a_n)[/Tex]

= (a₁-a₁) (a₁-a₂)………… (a₁-a_n)

[Tex]\lim_{x \to a_1} f(x) [/Tex]= 0

Now, let’s calculate for[Tex]\lim_{x \to a} f(x)[/Tex]

[Tex]\lim_{x \to a} f(x) = \lim_{x \to a} (x-a_1) (x-a_2)………… (x-a_n)[/Tex]

=[Tex]\lim_{x \to a} (x-a_1) \lim_{x \to a}(x-a_2)………… \lim_{x \to a}(x-a_n)[/Tex]

= (a-a₁) (a-a₂)………… (a-a_n)

[Tex]\lim_{x \to a} f(x) [/Tex]= (a-a₁) (a-a₂)………… (a-a_n)

Question 30: If[Tex]f(x)= \begin{cases} |x|+1, \hspace{0.2cm}x<0\\ 0,\hspace{0.2cm}x=0\\ |x|-1,\hspace{0.2cm}x>0 \end{cases}[/Tex]

For what value (s) of a does[Tex]\lim_{x \to a} f(x) [/Tex]exists?

Solution:

Here,

As, we know that mod function works differently.

In |x-0|, |x|=x when x>0 and |x|=-x when x<0

Let’s check for three cases of a:

• When a=0

Let’s calculate, the limits when x⇢0

Left limit =[Tex]\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (|x|+1)\\= -x+1\\=1[/Tex]

Right limit =[Tex]\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (|x|-1)\\= x-1\\=-1[/Tex]

As,[Tex]\lim_{x \to 0^-} f(x) \neq \lim_{x \to 0^+} f(x)[/Tex]

Hence, limit does not exists when x⇢0.

• When a>0

Let’s take a=2, for reference

Let’s calculate, the limits when x⇢2

Left limit =[Tex]\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (|x|-1)\\= x-1\\=2-1\\=1[/Tex]

Right limit =[Tex]\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (|x|-1)\\= x-1\\=2-1\\=1[/Tex]

As,[Tex]\lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x)[/Tex]

Hence, limit exists when x⇢2.

• When a<0

Let’s take a=-2, for reference

Let’s calculate, the limits when x⇢ -2

Left limit =[Tex]\lim_{x \to -2^-} f(x) = \lim_{x \to -2^-} (|x|+1)\\= x+1\\=-2+1\\=-1[/Tex]

Right limit =[Tex]\lim_{x \to -2^+} f(x) = \lim_{x \to -2^+} (|x|+1)\\= x+1\\=-2+1\\=-1[/Tex]

As,[Tex]\lim_{x \to -2^-} f(x) = \lim_{x \to -2^+} f(x)[/Tex]

Hence, limit exists when x⇢ -2.

Question 31: If the function f(x) satisfies[Tex]\lim_{x \to 1} \frac{f(x)-2}{x^2-1} = \pi [/Tex], evaluate[Tex]\lim_{x \to 1} f(x)[/Tex]

Solution:

Here, as it is given

[Tex]\lim_{x \to 1} \frac{f(x)-2}{x^2-1} = \pi[/Tex]

[Tex]\frac{\lim_{x \to 1} f(x)-2}{\lim_{x \to 1} x^2-1} = \pi[/Tex]

[Tex]\lim_{x \to 1} (f(x)-2) = \pi (\lim_{x \to 1} x^2-1)[/Tex]

Put x = 1 in RHS, we get

[Tex]\lim_{x \to 1} (f(x)-2) = \pi (\lim_{x \to 1} (1^2-1))[/Tex]

[Tex]\lim_{x \to 1} (f(x)-2) = 0[/Tex]

[Tex]\lim_{x \to 1} f(x)-\lim_{x \to 1} 2= 0[/Tex]

[Tex]\lim_{x \to 1} f(x) [/Tex]= 2

Hence proved!

Question 32: If[Tex]f(x)= \begin{cases} mx^2+n, \hspace{0.2cm}x<0\\ nx+m,\hspace{0.2cm},0\leq x\leq 1\\ nx^3+m, \hspace{0.2cm}x>1 \end{cases} [/Tex]. For what integers m and n does both[Tex]\lim_{x \to 0} f(x) [/Tex]and[Tex]\lim_{x \to 1} f(x) [/Tex]exists?

Solution:

Let’s calculate, the limits when x⇢0

Here,

Left limit =[Tex]\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (mx^2+n)\\ = (m(0)^2+n)\\ =n[/Tex]

Right limit =[Tex]\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (nx+m)\\ = (n(0)+m)\\ =m[/Tex]

Hence,

[Tex]\lim_{x \to 0^-} f(x)=\lim_{x \to 0^+} f(x) [/Tex], then limit exists

m = n

Now, let’s calculate, the limits when x⇢1

Here,

Left limit =[Tex]\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (nx+m)\\= (n(1)+m)\\=n+m[/Tex]

Right limit =[Tex]\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (nx^3+m)\\= (n(1)^3+m)\\=n+m[/Tex]

Hence,[Tex]\lim_{x \to 1^-} f(x)=\lim_{x \to 1^+} f(x) = m+n [/Tex], then limit exists.