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

Evaluate the following limits in Exercises 1 to 22.

Question 1: \lim_{x \to 3} x+3

Solution:

In \lim_{x \to 3} x+3, as x⇢3 

Put x = 3, we get

\lim_{x \to 3} x+3 = 3+3 

= 6

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

Solution:

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

Put x = π, we get

\lim_{x \to \pi} (x-\frac{22}{7}) = (π-\frac{22}{7}) 

= (π-\frac{22}{7})

Question 3: \lim_{r \to 1} \pi r^2

Solution:

In \lim_{r \to 1} \pi r^2, as r⇢1

Put r = 1, we get

\lim_{r \to 1} \pi r^2 = \pi (1)^2 

= π

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

Solution:

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

Put x = 4, we get

\lim_{x \to 4} (\frac{4x+3}{x-2}) = \frac{4(4)+3}{4-2} 

= \frac{19}{2}

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

Solution:

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

Put x = -1, we get

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

= \frac{1-1+1}{-2}

= \frac{-1}{2}

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

Solution:

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

Put x = 0, we get

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

As, this limit becomes undefined

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

\mathbf{\lim_{x \to a} \frac{x^n-a^n}{x-a} = na^{n-1}}

As, x⇢0 ⇒ p⇢1

\lim_{p \to 1} \frac{(p)^5-1}{p-1} = \lim_{p \to 1} \frac{(p)^5-1^5}{p-1}

Here, n=5 and a = 1.

\lim_{p \to 1} \frac{(p)^5-1}{p-1} = 5(1)^{5-1} 

= 5(1)⁴ 

= 5

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

Solution:

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

Put x = 2, we get

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

As, this limit becomes undefined

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

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

= \lim_{x \to 2} \frac{(3x+5)(x-2)}{(x+2)(x-2)}

Cancelling (x-2), we have

= \lim_{x \to 2} \frac{(3x+5)}{(x+2)}

Put x = 2, we get

\lim_{x \to 2} \frac{(3x+5)}{(x+2)} = \frac{(3(2)+5)}{(2+2)} 

= \frac{11}{4} 

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

Solution:

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

Put x = 3, we get

\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}

As, this limit becomes undefined

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

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

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

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

Cancelling (x-3), we have

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

Put x = 3, we get

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

= \frac{9\times 18}{7} 

= \frac{108}{7} 

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

Solution:

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

Put x = 0, we get

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

= b

Question 10: \lim_{z \to 1} \frac{z^{\frac{1}{3}}-1}{z^{\frac{1}{6}}-1}

Solution:

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

Put z = 1, we get

\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}

Let's take z^{\frac{1}{6}}   = p and z^{\frac{1}{3}}   = p²,

As, z⇢1 ⇒ p⇢1

= \lim_{p \to 1} \frac{p^2-1}{p-1}

Now, let's Factorise the numerator, we get

= \lim_{p \to 1} \frac{(p-1)(p+1)}{p-1}

Cancelling (p-1), we have

= \lim_{p \to 1} (p+1)

Put p = 1, we get

\lim_{p \to 1} (p+1) = 1+1

= 2

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

Solution:

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

Put x = 1, we get

\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}

= \frac{a+b+c}{c+b+a}

= 1 (As it is given a+b+c≠0)

Question 12: \lim_{x \to -2} \frac{\frac{1}{x} + \frac{1}{2}}{x+2}

Solution:

In \lim_{x \to -2} \frac{\frac{1}{x} + \frac{1}{2}}{x+2}, as x⇢-2

Firstly, lets simplify the equation

\frac{\frac{1}{x} + \frac{1}{2}}{x+2} = \frac{\frac{2+x}{2x}}{x+2}

Cancelling (x+2),we get

\lim_{x \to -2} \frac{1}{2x}

Put x = -2, we get

\lim_{x \to -2} \frac{1}{2x} = \frac{1}{2(-2)}

= \frac{-1}{4}

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

Solution:

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

Put x = 0, we get

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

As, this limit becomes undefined

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

\mathbf{\lim_{x \to 0} \frac{sin \hspace{0.1cm}x}{x} = 1}

Hence, we have

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

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

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

As x⇢0, then ax⇢0

= \frac{a}{b} \lim_{ax \to 0} \frac{sin\hspace{0.1cm}ax}{ax}

By using the theorem, we get

 = \frac{a}{b} . 1

= \frac{a}{b}

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

Solution:

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

Put x = 0, we get

\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}

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.

\mathbf{\lim_{x \to 0} \frac{sin \hspace{0.1cm}x}{x} = 1}

Hence, we have

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

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

= \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}}

By using the theorem, we get

= \frac{a}{b} . 1 .1

= \frac{a}{b}

Question 15: \lim_{x \to \pi} \frac{sin(\pi-x)}{\pi(\pi-x)}

Solution:

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

Put x = π, we get

\lim_{x \to \pi} \frac{sin(\pi-x)}{\pi(\pi-x)} = \frac{sin(\pi-\pi)}{\pi(\pi-\pi)} = \frac{0}{0}

As, this limit becomes undefined

Now, let's take π-x=p

As, x⇢π ⇒ p⇢0

\mathbf{\lim_{x \to 0} \frac{sin \hspace{0.1cm}x}{x} = 1}

= \lim_{p \to 0} \frac{sin p}{p\pi}

= \frac{1}{\pi} \lim_{p \to 0} \frac{sin\hspace{0.1cm} p}{p}

By using the theorem, we get

= 1. \frac{1}{\pi}

= \frac{1}{\pi}

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

Solution:

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

Put x = 0, we get

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

= \frac{1}{\pi}

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

Solution:

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

As we know, cos 2θ = 1-2sin²θ

Substituting the values, we get

\lim_{x \to 0} \frac{1-2sin^2x-1}{1-2sin^2(\frac{x}{2})-1}

=\lim_{x \to 0} \frac{sin^2x}{sin^2(\frac{x}{2})}

Put x = 0, we get

\lim_{x \to 0} \frac{sin^2x}{sin^2(\frac{x}{2})} = \frac{0}{0}

As, this limit becomes undefined

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

\mathbf{\lim_{x \to 0} \frac{sin x}{x} = 1}

Hence, we have

\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}}

=\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}

=\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}

By using the theorem, we get

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

=\lim_{x \to 0}\frac{x^2}{(\frac{x^2}{4})}

=\lim_{x \to 0}(4)

= 4

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

Solution:

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

Put x = 0, we get

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

As, this limit becomes undefined

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

\mathbf{\lim_{x \to 0} \frac{sin x}{x} = 1}

Hence, we have

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

=\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)

By using the theorem, we get

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

=\frac{1}{b}\times a

Putting x=0, we have

=\frac{a}{b}

Question 19:\lim_{x \to 0} x sec\hspace{0.1cm}x

Solution:

In\lim_{x \to 0} x sec\hspace{0.1cm}x , as x⇢0

Put x = 0, we get

\lim_{x \to 0} x sec\hspace{0.1cm}x = 0 ×1

= 0

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

Solution:

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

Put x = 0, we get

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

As, this limit becomes undefined

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

\mathbf{\lim_{x \to 0} \frac{sin x}{x} = 1}

Hence, we can write the equation as follows:

\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}}

=\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}

By using the theorem, we get

=\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}

=\frac{\lim_{x \to 0}ax+\lim_{x \to 0}bx}{\lim_{x \to 0}ax+\lim_{x \to 0} bx}

=\lim_{x \to 0} \frac{ax+bx}{ax+bx}

=\lim_{x \to 0} 1

Putting x=0, we have

= 1

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

Solution:

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

By simplification, we get

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

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

Put x = 0, we get

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

As, this limit becomes undefined

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

\mathbf{\lim_{x \to 0} \frac{sin x}{x} = 1}

By using the trigonometric identities,

cos 2θ = 1-2sin²θ

sin 2θ = 2 sinθ cosθ

Hence, we can write the equation as follows:

\lim_{x \to 0} (\frac{2sin^2(\frac{x}{2})}{2 sin(\frac{x}{2})cos(\frac{x}{2})})

=\lim_{x \to 0} (\frac{sin(\frac{x}{2})}{cos(\frac{x}{2})})

=\lim_{x \to 0} tan(\frac{x}{2})

Putting x=0, we have

= 0

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

Solution:

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

Put x =\frac{\pi}{2} , we get

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

As, this limit becomes undefined

Now, let's simplify the equation :

Let's takex-\frac{\pi}{2}=p

As, x⇢\frac{\pi}{2} ⇒ p⇢0

Hence, we can write the equation as follows:

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

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

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

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

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

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

\mathbf{\lim_{x \to 0} \frac{sin x}{x} = 1}

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

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

As p⇢0, then 2p⇢0

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

Using the theorem and putting p=0, we have

= 2×1×1

= 2

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

Solution:

Let's calculate, the limits when x⇢0

Here,

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

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

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

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

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

Here,

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

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

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

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

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

Solution:

Let's calculate, the limits when x⇢1

Here,

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

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

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

Hence, limit does not exists when x⇢1.

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

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 =\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{|x|}{x}\\= \frac{-x}{x}\\=-1

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

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

Hence, limit does not exists when x⇢0.

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

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 =\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{x}{|x|}\\= \frac{x}{-x}\\=-1

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

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

Hence, limit does not exists when x⇢0.

Question 27: Find\lim_{x \to 5} f(x) , 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 =\lim_{x \to 5^-} f(x) = \lim_{x \to 5^-} |x|-5\\= x-5\\=5-5\\=0

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

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

Question 28: Supposef(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} and if\lim_{x \to 1} f(x) = f(1) what are possible values of a and b?

Solution:

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

Let's calculate, the limits when x⇢1

Here,

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

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

Limit value f(1) = 4

So, as limit exists then it should satisfy

\lim_{x \to 1^-} f(x)= \lim_{x \to 1} f(x)=\lim_{x \to 1^+} f(x) = f(1) = 4

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\lim_{x \to a_1} f(x) ? For some a ≠ a₁, a₂, ..., an, compute\lim_{x \to a} f(x).

Solution:

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

Then,\lim_{x \to a_1} f(x) = \lim_{x \to a_1} (x-a_1) (x-a_2)............ (x-a_n)

=\lim_{x \to a_1} (x-a1) \lim_{x \to a_1}(x-a_2)............ \lim_{x \to a_1}(x-a_n)

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

\lim_{x \to a_1} f(x) = 0

Now, let's calculate for\lim_{x \to a} f(x)

\lim_{x \to a} f(x) = \lim_{x \to a} (x-a_1) (x-a_2)............ (x-a_n)

=\lim_{x \to a} (x-a_1) \lim_{x \to a}(x-a_2)............ \lim_{x \to a}(x-a_n)

= (a-a₁) (a-a₂)............ (a-a_n)

\lim_{x \to a} f(x) = (a-a₁) (a-a₂)............ (a-a_n)

Question 30: Iff(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}

For what value (s) of a does\lim_{x \to a} f(x) 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 =\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (|x|+1)\\= -x+1\\=1

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

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

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 =\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (|x|-1)\\= x-1\\=2-1\\=1

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

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

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 =\lim_{x \to -2^-} f(x) = \lim_{x \to -2^-} (|x|+1)\\= x+1\\=-2+1\\=-1

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

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

Hence, limit exists when x⇢ -2.

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

Solution:

Here, as it is given

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

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

\lim_{x \to 1} (f(x)-2) = \pi (\lim_{x \to 1} x^2-1)

Put x = 1 in RHS, we get

\lim_{x \to 1} (f(x)-2) = \pi (\lim_{x \to 1} (1^2-1))

\lim_{x \to 1} (f(x)-2) = 0

\lim_{x \to 1} f(x)-\lim_{x \to 1} 2= 0

\lim_{x \to 1} f(x) = 2

Hence proved!

Question 32: Iff(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} . For what integers m and n does both\lim_{x \to 0} f(x) and\lim_{x \to 1} f(x) exists?

Solution:

Let's calculate, the limits when x⇢0

Here,

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

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

Hence,

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

m = n

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

Here,

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

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

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