类11 NCERT解决方案-第13章极限和导数–练习13.1 |套装1

在练习1至22中评估以下限制。

问题1： $\lim_{x \to 3} x+3$

解决方案：

In $\lim_{x \to 3} x+3$, as x⇢3 

Put x = 3, we get

$\lim_{x \to 3} x+3$ = 3+3 

= 6

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

解决方案：

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

Put x = π, we get

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

= $(π-\frac{22}{7})$

问题3： $\lim_{r \to 1} \pi r^2$

解决方案：

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

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

解决方案：

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

Put x = 4, we get

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

= $\frac{19}{2}$

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

解决方案：

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

Put x = -1, we get

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

= $\frac{1-1+1}{-2}$

= $\frac{-1}{2}$

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

解决方案：

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

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

解决方案：

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

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

= $\frac{11}{4}$

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

解决方案：

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

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

= $\frac{9\times 18}{7}$

= $\frac{108}{7}$

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

解决方案：

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

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

解决方案：

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

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

解决方案：

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)

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

解决方案：

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

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

= $\frac{-1}{4}$

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

解决方案：

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

 = $\frac{a}{b} . 1$

= $\frac{a}{b}$

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

解决方案：

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

= $\frac{a}{b} . 1 .1$

= $\frac{a}{b}$

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

解决方案：

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

= $1. \frac{1}{\pi}$

= $\frac{1}{\pi}$

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

解决方案：

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

Put x = 0, we get

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

= $\frac{1}{\pi}$