类11 NCERT解决方案-第13章极限和导数–练习13.1 |套装2

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

解决方案：

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

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

解决方案：

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

=$\frac{a}{b}$

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

解决方案：

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

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

解决方案：

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

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

解决方案：

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

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

解决方案：

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 take$x-\frac{\pi}{2}=p$

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

问题23：查找$\lim_{x \to 0} f(x)$和$\lim_{x \to 1} f(x)$ ， 在哪里$f(x)= \begin{cases} 2x+3, \hspace{0.2cm}x\leq0\\ 3(x+1),\hspace{0.2cm}x>0 \end{cases}$

解决方案：

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

问题24：查找$\lim_{x \to 1} f(x)$ ， 在哪里$f(x)= \begin{cases} x^2-1, \hspace{0.2cm}x\leq1\\ -x^2-1,\hspace{0.2cm}x>1 \end{cases}$

解决方案：

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.

问题25：评估$\lim_{x \to 0} f(x)$ ， 在哪里$f(x)= \begin{cases} \frac{|x|}{x}, \hspace{0.2cm}x\neq0\\ 0,\hspace{0.2cm}x=0 \end{cases}$

解决方案：

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.

问题26：查找$\lim_{x \to 0} f(x)$ ， 在哪里$f(x)= \begin{cases} \frac{x}{|x|}, \hspace{0.2cm}x\neq0\\ 0,\hspace{0.2cm}x=0 \end{cases}$

解决方案：

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.

问题27：查找$\lim_{x \to 5} f(x)$ ，其中f(x)= | x | -5。

解决方案：

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

问题28：假设$f(x)= \begin{cases} a+bx, \hspace{0.2cm}x<1\\ 4,\hspace{0.2cm}x=1\\ b-ax,\hspace{0.2cm}x>1 \end{cases}$而如果$\lim_{x \to 1} f(x) = f(1)$ a和b的可能值是多少?

解决方案：

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

问题29：让a ₁ ，a ₂ ，。 。 。， _n是固定的实数并定义一个函数

f(x)=(xa ₁ )(xa ₂ )…………(x-an)。

什么是$\lim_{x \to a_1} f(x)$ ?对于一些≠a ₁ ，a ₂ ，…，an计算$\lim_{x \to a} f(x)$ 。

解决方案：

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)

问题30：如果$f(x)= \begin{cases} |x|+1, \hspace{0.2cm}x<0\\ 0,\hspace{0.2cm}x=0\\ |x|-1,\hspace{0.2cm}x>0 \end{cases}$

对于a的什么值$\lim_{x \to a} f(x)$存在吗?

解决方案：

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.

问题31：如果函数f(x)满足$\lim_{x \to 1} \frac{f(x)-2}{x^2-1} = \pi$ ， 评估$\lim_{x \to 1} f(x)$

解决方案：

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!

问题32：如果$f(x)= \begin{cases} mx^2+n, \hspace{0.2cm}x<0\\ nx+m,\hspace{0.2cm},0\leq x\leq 1\\ nx^3+m, \hspace{0.2cm}x>1 \end{cases}$ 。对于整数m和n两者$\lim_{x \to 0} f(x)$和$\lim_{x \to 1} f(x)$存在吗?

解决方案：

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.