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⇢3Put 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⇢πPut 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⇢1Put 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⇢4Put 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⇢-1Put 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⇢0Put 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⇢2Put 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⇢3Put 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⇢0Put 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⇢1Put 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 andz^{\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⇢1Put 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⇢-2Firstly, 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⇢0Put 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⇢0Put 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⇢πPut 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⇢0Put 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}