第12类NCERT解决方案-数学第I部分–第5章连续性和可微性–练习5.1 |套装2

问题18：对于λ的哪个值，该函数定义为

$f(x)= \begin{cases} \lambda(x^2-2x), \hspace{0.2cm}x \leq0\\ 4x+1,\hspace{0.2cm}x>0 \end{cases}$

连续于x = 0?那么在x = 1时的连续性又如何呢?

解决方案：

To be continuous function, f(x) should satisfy the following at x = 0:

$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0)$

Continuity at x = 0,

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

= λ(0²– 2(0)) = 0

Right limit = $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (4x+1)$

= λ4(0) + 1 = 1

Function value at x = 0, f(0) = $\lambda(0^2-2(0)) = 0$

As, 0 = 1 cannot be possible

Hence, for no value of λ, f(x) is continuous.

But here, $\lim_{x \to 0^-} f(x) \neq \lim_{x \to 0^+} f(x)$

Continuity at x = 1,

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

= (4(1) + 1) = 5

Right limit = $\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (4x+1)$

= 4(1) + 1 = 5

Function value at x = 1, f(1) = 4(1) + 1 = 5

As, $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1) = 5$

Hence, the function is continuous at x = 1 for any value of λ.

问题19：证明由g(x)= x – [x]定义的函数在所有积分点上都是不连续的。此处[x]表示小于或等于x的最大整数。

解决方案：

[x] is greatest integer function which is defined in all integral points, e.g.

[2.5] = 2

[-1.96] = -2

x-[x] gives the fractional part of x.

e.g: 2.5 – 2 = 0.5

c be an integer

Let’s check the continuity at x = c,

Left limit = $\lim_{x \to c^-} f(x) = \lim_{x \to c^-} (x-[x])$

= (c – (c – 1)) = 1

Right limit = $\lim_{x \to c^+} f(x) = \lim_{x \to c^+} (x-[x])$

= (c – c) = 0

Function value at x = c, f(c) = c – = c – c = 0

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

Hence, the function is discontinuous at integral.

c be not an integer

Let’s check the continuity at x = c,

Left limit = $\lim_{x \to c^-} f(x) = \lim_{x \to c^-} (x-[x])$

= (c – (c – 1)) = 1

Right limit = $\lim_{x \to c^+} f(x) = \lim_{x \to c^+} (x-[x])$

= (c – (c – 1)) = 1

Function value at x = c, f(c) = c – = c – (c – 1) = 1

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

Hence, the function is continuous at non-integrals part.

问题20.由f(x)= x ² – sin x + 5定义的函数在x =π处连续吗?

解决方案：

Let’s check the continuity at x = π,

f(x) = x² – sin x + 5

Let’s substitute, x = π+h

When x⇢π, Continuity at x = π

Left limit = $\lim_{x \to \pi^-} f(x) = \lim_{x \to \pi^-} (x^2 - sin \hspace{0.1cm}x + 5)$

= (π² – sinπ + 5) = π² + 5

Right limit = $\lim_{x \to \pi^+} f(x) = \lim_{x \to \pi^+}(x^2 - sin \hspace{0.1cm}x + 5)$

= (π² – sinπ + 5) = π² + 5

Function value at x = π, f(π) = π² – sin π + 5 = π² + 5

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

Hence, the function is continuous at x = π .

问题21.讨论以下功能的连续性：

(a)f(x)= sin x + cos x

解决方案：

Here, 

f(x) = sin x + cos x

Let’s take, x = c + h

When x⇢c then h⇢0

$\lim_{x \to c} f(x) = \lim_{h \to 0} f(c+h)$

So, 

$\lim_{h \to 0} f(c+h) = \lim_{h \to 0}$(sin(c + h) + cos(c + h))

Using the trigonometric identities, we get

sin(A + B) = sin A cos B + cos A sin B

cos(A + B) =  cos A cos B – sin A sin B

$\lim_{h \to 0} f(c+h) = \lim_{h \to 0}$((sinc cosh + cosc sinh) + (cosc cosh − sinc sinh))

$\lim_{h \to 0} f(c+h)$= ((sinc cos0 + cosc sin0) + (cosc cos0 − sinc sin0))

cos 0 = 1 and sin 0 = 0

$\lim_{h \to 0} f(c+h)$ = (sinc + cosc) = f(c)

Function value at x = c, f(c) = sinc + cosc

As, $\lim_{x \to c} f(x)$= f(c) = sinc + cosc

Hence, the function is continuous at x = c.

(b)f(x)= sin x – cos x

解决方案：

Here,

f(x) = sin x – cos x

Let’s take, x = c+h

When x⇢c then h⇢0

$\lim_{x \to c} f(x) = \lim_{h \to 0} f(c+h)$

So,

$\lim_{h \to 0} f(c+h) = \lim_{h \to 0}$(sin(c + h) − cos(c + h))

Using the trigonometric identities, we get

sin(A + B) = sin A cos B + cos A sin B

cos(A + B) =  cos A cos B – sin A sin B

$\lim_{h \to 0} f(c+h) = \lim_{h \to 0}$((sinc cosh + cosc sinh) − (cosc cosh − sinc sinh))

$\lim_{h \to 0} f(c+h)$ = ((sinc cos0 + cosc sin0) − (cosc cos0 − sinc sin0))

cos 0 = 1 and sin 0 = 0

$\lim_{h \to 0} f(c+h)$ = (sinc − cosc) = f(c)

Function value at x = c, f(c) = sinc − cosc

As, $\lim_{x \to c} f(x)$ = f(c) = sinc − cosc

Hence, the function is continuous at x = c.

(c)f(x)= sin x。 cos x

解决方案：

Here,

f(x) = sin x + cos x

Let’s take, x = c+h

When x⇢c then h⇢0

$\lim_{x \to c} f(x) = \lim_{h \to 0} f(c+h)$

So,

$\lim_{h \to 0} f(c+h) = \lim_{h \to 0}$sin(c + h) × cos(c + h))

Using the trigonometric identities, we get

sin(A + B) = sin A cos B + cos A sin B

cos(A + B) =  cos A cos B – sin A sin B

$\lim_{h \to 0} f(c+h) = \lim_{h \to 0}$((sinc cosh + cosc sinh) × (cosc cosh − sinc sinh))

$\lim_{h \to 0} f(c+h)$= ((sinc cos0 + cosc sin0) × (cosc cos0 − sinc sin0))

cos 0 = 1 and sin 0 = 0

$\lim_{h \to 0} f(c+h)$ = (sinc × cosc) = f(c)

Function value at x = c, f(c) = sinc × cosc

As, $\lim_{x \to c} f(x)$= f(c) = sinc × cosc

Hence, the function is continuous at x = c.

问题22.讨论余弦，割线，割线和切线函数的连续性。

解决方案：

Continuity of cosine

Here,

f(x) = cos x

Let’s take, x = c+h

When x⇢c then h⇢0

$\lim_{x \to c} f(x) = \lim_{h \to 0} f(c+h)$

So,

$\lim_{h \to 0} f(c+h) = \lim_{h \to 0} (cos\hspace{0.1cm} (c+h))$

Using the trigonometric identities, we get

cos(A + B) =  cos A cos B – sin A sin B

$\lim_{h \to 0} f(c+h) = \lim_{h \to 0}$(cosc cosh − sinc sinh)

$\lim_{h \to 0} f(c+h)$ = (cosc cos0 − sinc sin0)

cos 0 = 1 and sin 0 = 0

$\lim_{h \to 0} f(c+h)$ = (cosc) = f(c)

Function value at x = c, f(c) = (cosc)

As, $\lim_{x \to c} f(x)$ = f(c) = (cosc)

Hence, the cosine function is continuous at x = c.

Continuity of cosecant

Here,

f(x) = cosec x = $\frac{1}{sin \hspace{0.1cm}x}$

Domain of cosec is R – {nπ}, n ∈ Interger

Let’s take, x = c + h

When x⇢c then h⇢0

$\lim_{x \to c} f(x) = \lim_{h \to 0} f(c+h)$

So,

$\lim_{h \to 0} f(c+h) = \lim_{h \to 0} (\frac{1}{sin \hspace{0.1cm}(c+h)})$

Using the trigonometric identities, we get

sin(A + B) = sin A cos B + cos A sin B

$\lim_{h \to 0} f(c+h) = \lim_{h \to 0} (\frac{1}{sin\hspace{0.1cm} c\hspace{0.1cm} cos\hspace{0.1cm} h+cos\hspace{0.1cm} c\hspace{0.1cm} sin\hspace{0.1cm} h})\\ \lim_{h \to 0} f(c+h) = \lim_{h \to 0} (\frac{1}{sin\hspace{0.1cm} c\hspace{0.1cm} cos\hspace{0.1cm} 0+cos\hspace{0.1cm} c\hspace{0.1cm} sin\hspace{0.1cm} 0})$

cos 0 = 1 and sin 0 = 0

$\lim_{h \to 0} f(c+h) = \lim_{h \to 0} (\frac{1}{sin\hspace{0.1cm} c})$

Function value at x = c, f(c) = $\frac{1}{sin\hspace{0.1cm} c}$

As, $\lim_{x \to c} f(x) = f(c) = \frac{1}{sin\hspace{0.1cm} c}$

Hence, the cosecant function is continuous at x = c.

Continuity of secant

Here,

f(x) = sec x = $\frac{1}{cos \hspace{0.1cm}x}$

Let’s take, x = c + h

When x⇢c then h⇢0

$\lim_{x \to c} f(x) = \lim_{h \to 0} f(c+h)$

So,

$\lim_{h \to 0} f(c+h) = \lim_{h \to 0} (\frac{1}{cos \hspace{0.1cm}(c+h)})$

Using the trigonometric identities, we get

cos(A + B) =  cos A cos B – sin A sin B

$\lim_{h \to 0} f(c+h) = \lim_{h \to 0} (\frac{1}{cos\hspace{0.1cm} c\hspace{0.1cm} cos\hspace{0.1cm} h-sin\hspace{0.1cm} c\hspace{0.1cm} sin\hspace{0.1cm} h})$

$\lim_{h \to 0} f(c+h) = \lim_{h \to 0} (\frac{1}{cos\hspace{0.1cm} c\hspace{0.1cm} cos\hspace{0.1cm} 0-sin\hspace{0.1cm} c\hspace{0.1cm} sin\hspace{0.1cm} 0})$

cos 0 = 1 and sin 0 = 0

$\lim_{h \to 0} f(c+h) = \lim_{h \to 0} (\frac{1}{cos\hspace{0.1cm} c})$

Function value at x = c, f(c) = $\frac{1}{cos\hspace{0.1cm} c}$

As, $\lim_{x \to c} f(x) = f(c) = \frac{1}{cos\hspace{0.1cm} c}$

Hence, the secant function is continuous at x = c.

Continuity of cotangent

Here,

f(x) = cot x = $\frac{cos \hspace{0.1cm}x}{sin \hspace{0.1cm}x}$

Let’s take, x = c+h

When x⇢c then h⇢0

$\lim_{x \to c} f(x) = \lim_{h \to 0} f(c+h)$

So,

$\lim_{h \to 0} f(c+h) = \lim_{h \to 0} (\frac{cos \hspace{0.1cm}(c+h)}{sin \hspace{0.1cm}(c+h)})$

Using the trigonometric identities, we get

sin(A + B) = sin A cos B + cos A sin B

cos(A + B) =  cos A cos B – sin A sin B

$\lim_{h \to 0} f(c+h) = \lim_{h \to 0} (\frac{cos\hspace{0.1cm} c\hspace{0.1cm} cos\hspace{0.1cm} h-sin\hspace{0.1cm} c\hspace{0.1cm} sin\hspace{0.1cm} h}{sin\hspace{0.1cm} c\hspace{0.1cm} cos\hspace{0.1cm} h+cos\hspace{0.1cm} c\hspace{0.1cm} sin\hspace{0.1cm} h})$

$\lim_{h \to 0} f(c+h) = \lim_{h \to 0} (\frac{cos\hspace{0.1cm} c\hspace{0.1cm} cos\hspace{0.1cm} 0-sin\hspace{0.1cm} c\hspace{0.1cm} sin\hspace{0.1cm} 0}{sin\hspace{0.1cm} c\hspace{0.1cm} cos\hspace{0.1cm} 0+cos\hspace{0.1cm} c\hspace{0.1cm} sin\hspace{0.1cm} 0})$

cos 0 = 1 and sin 0 = 0

$\lim_{h \to 0} f(c+h) = \lim_{h \to 0} (\frac{cos\hspace{0.1cm} c}{sin\hspace{0.1cm} c})$

$\lim_{h \to 0} f(c+h) = (\frac{cos\hspace{0.1cm} c}{sin\hspace{0.1cm} c})$

Function value at x = c, f(c) = $\frac{cos\hspace{0.1cm} c}{sin\hspace{0.1cm} c}$

As, $\lim_{x \to c} f(x) = f(c) = \frac{cos\hspace{0.1cm} c}{sin\hspace{0.1cm} c}$

Hence, the cotangent function is continuous at x = c.

问题23：找出f的所有不连续点，其中

$f(x)= \begin{cases} \frac{sin \hspace{0.1cm}x}{x}, \hspace{0.2cm}x <0\\ x+1,\hspace{0.2cm}x\geq0 \end{cases}$

解决方案：

Here,

From the two continuous functions g and h, we get

$\frac{g(x)}{h(x)}$= continuous when h(x) ≠ 0

For x < 0, f(x) = $\frac{sin \hspace{0.1cm}x}{x}$, is continuous

Hence, f(x) is continuous x ∈ (-∞, 0)

Now, For x ≥ 0, f(x) = x + 1, which is a polynomial

As polynomial are continuous, hence f(x) is continuous x ∈ (0, ∞)

So now, as f(x) is continuous in x ∈ (-∞, 0) U (0, ∞)= R – {0}

Let’s check the continuity at x = 0,

Left limit = $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (\frac{sin \hspace{0.1cm}x}{x})\\= 1$

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

Function value at x = 0, f(0) = 0 + 1 = 1

As, $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) = 1$

Hence, the function is continuous at x = 0.

Hence, the function is continuous for any real number.

问题24.确定f是否定义为

$f(x)= \begin{cases} x^2sin\frac{1}{x}, \hspace{0.2cm}x \neq0\\ 0,\hspace{0.2cm}x=0 \end{cases}$

是连续函数吗?

解决方案：

Here, as it is given that

For x = 0, f(x) = 0, which is a constant

As constant are continuous, hence f(x) is continuous x ∈ = R – {0}

Let’s check the continuity at x = 0,

As, we know range of sin function is [-1,1]. So, -1 ≤ $sin(\frac{1}{x})$≤ 1 which is a finite number.

Limit = $\lim_{x \to 0} f(x) = \lim_{x \to 0} (x^2 sin(\frac{1}{x}))$

= (0² ×(finite number)) = 0

Function value at x = 0, f(0) = 0

As, $\lim_{x \to 0} f(x) = f(0).$

Hence, the function is continuous for any real number.

问题25.检查f的连续性，其中f定义为

$f(x)= \begin{cases} sin\hspace{0.1cm}x-cos\hspace{0.1cm}x, \hspace{0.2cm}x\neq0\\ -1,\hspace{0.2cm}x=0 \end{cases}$

解决方案：

Continuity at x = 0,

Left limit = $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (sin\hspace{0.1cm}x-cos\hspace{0.1cm}x)$

= (sin0 − cos0) = 0 − 1 = −1

Right limit = $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (sin\hspace{0.1cm}x-cos\hspace{0.1cm}x)$

= (sin0 − cos0) = 0 − 1 = −1

Function value at x = 0, f(0) = sin 0 – cos 0 = 0 – 1 = -1

As, $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) = -1$

Hence, the function is continuous at x = 0.

Continuity at x = c (real number c≠0),

Left limit = $\lim_{x \to c^-} f(x) = \lim_{x \to c^-} (sin\hspace{0.1cm}x-cos\hspace{0.1cm}x)$

= (sinc − cosc)

Right limit = $\lim_{x \to c^+} f(x) = \lim_{x \to c^+} (sin\hspace{0.1cm}x-cos\hspace{0.1cm}x)$

= (sinc − cosc)

Function value at x = c, f(c) = sin c – cos c

As, $\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c) = (sin\hspace{0.1cm}c-cos\hspace{0.1cm}c)$

So concluding the results, we get

The function f(x) is continuous at any real number.

找到k的值，以使函数f在练习26至29中的指定点处连续。

问题26。 $f(x)= \begin{cases} \frac{k\hspace{0.1cm}cos\hspace{0.1cm}x}{\pi-2x}, \hspace{0.2cm}x\neq\frac{\pi}{2}\\ 3,x=\frac{\pi}{2} \end{cases} \hspace{0.1cm}\hspace{0.1cm}$在x = π/ 2。

解决方案：

Continuity at x = π/2

Let’s take x = $\frac{\pi}{2}+h$

When x⇢π/2 then h⇢0

Substituting x = $\frac{\pi}{2}$+h, we get

cos(A + B) = cos A cos B – sin A sin B

Limit = $\lim_{h \to 0} f(\frac{\pi}{2}+h) = \lim_{h \to 0} (\frac{k\hspace{0.1cm}cos(\frac{\pi}{2}+h)}{\pi-2(\frac{\pi}{2}+h)}\\= \lim_{h \to 0} (\frac{k(cos(\frac{\pi}{2})cos h-sin(\frac{\pi}{2})sinh)}{\pi-\pi-2h)}\\= \lim_{h \to 0} (\frac{k(0 \times cos h-1\times sinh)}{-2h)}\\= \lim_{h \to 0} (\frac{k(-sinh)}{-2h)}\\ = \frac{k}{2} \lim_{h \to 0} (\frac{(sinh)}{h)}\\ = \frac{k}{2}$

Function value at x = $\frac{\pi}{2}, f(\frac{\pi}{2})$= 3

As, $\lim_{x \to \frac{\pi}{2}} f(x) = f(\frac{\pi}{2})$should satisfy, for f(x) being continuous

k/2 = 3

k = 6

问题27。 $f(x)= \begin{cases} kx^2,x\leq2\\ 3,x>2 \end{cases} \hspace{0.1cm}\hspace{0.1cm}$在x = 2

解决方案：

Continuity at x = 2

Left limit = $\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (kx^2)$

= k(2)² = 4k

Right limit = $\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (3)\\= 3$

Function value at x = 2, f(2) = k(2)² = 4k

As, $\lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x)= f(2)$ should satisfy, for f(x) being continuous

4k = 3

k = 3/4

问题28。 $f(x)= \begin{cases} kx+1,x\leq\pi\\ cos \hspace{0.2cm}x,x>\pi \end{cases} \hspace{0.1cm}\hspace{0.1cm}$在x =π处

解决方案：

Continuity at x = π

Left limit = $\lim_{x \to \pi^-} f(x) = \lim_{x \to \pi^-} (kx+1)$

= k(π) + 1

Right limit = $\lim_{x \to \pi^+} f(x) = \lim_{x \to \pi^+} (cos x)$

= cos(π) = -1

Function value at x = π, f(π) = k(π) + 1

As, $\lim_{x \to \pi^-}f(x) = \lim_{x \to \pi^+} f(x)= f(\pi)$ should satisfy, for f(x) being continuous

kπ + 1 = -1

k = -2/π

问题29。 $f(x)= \begin{cases} kx+1,x\leq5\\ 3x-5,x>5 \end{cases} \hspace{0.1cm}\hspace{0.1cm}$在x = 5

解决方案：

Continuity at x = 5

Left limit = $\lim_{x \to 5^-} f(x) = \lim_{x \to 5^-} (kx+1)$

= k(5) + 1 = 5k + 1

Right limit = $\lim_{x \to 5^+} f(x) = \lim_{x \to 5^+} (3x-5)$

= 3(5) – 5 = 10

Function value at x = 5, f(5) = k(5) + 1 = 5k + 1

As, $\lim_{x \to 5^-} f(x) = \lim_{x \to 5^+} f(x)= f(5)$should satisfy, for f(x) being continuous

5k + 1 = 10

k = 9/5 

问题30.找到a和b的值，使得由定义的函数

$f(x)= \begin{cases} 5,x\leq2\\ ax+b,2<x<10\\ 21,x\geq10 \end{cases}$

是一个连续的函数

解决方案：

Continuity at x = 2

Left limit = $\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (5)\\= 5$

Right limit = $\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (ax+b)\\= 2a+b$

Function value at x = 2, f(2) = 5

As, $\lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x)= f(2)$should satisfy, for f(x) being continuous at x = 2

2a + b = 5 ……………………(1)

Continuity at x = 10

Left limit = $\lim_{x \to 10^-} f(x) = \lim_{x \to 10^-} (ax+b)$

= 10a + b

Right limit = $\lim_{x \to 10^+} f(x) = \lim_{x \to 10^+} (21)$

= 21

Function value at x = 10, f(10) = 21

As, $\lim_{x \to 10^-} f(x) = \lim_{x \to 10^+} f(x)= f(10)$should satisfy, for f(x) being continuous at x = 10

10a + b = 21 ……………………(2)

Solving the eq(1) and eq(2), we get

a = 2

b = 1

问题31.证明由f(x)= cos(x ² )定义的函数是一个连续函数

解决方案：

Let’s take

g(x) = cos x

h(x) = x²

g(h(x)) = cos (x²)

To prove g(h(x)) continuous, g(x) and h(x) should be continuous.

Continuity of g(x) = cos x

Let’s check the continuity at x = c

x = c + h

g(c + h) = cos (c + h)

When x⇢c then h⇢0

cos(A + B) =  cos A cos B – sin A sin B

Limit = $\lim_{h \to 0} g(c+h) = \lim_{h \to 0} (cos(c+h))\\ = \lim_{h \to 0}$(cosc cosh − sinc sinh)

= cosc cos0 − sinc sin0 = cosc

Function value at x = c, g(c) = cos c

As, $\lim_{x \to c} g(x) = g(c) = cos\hspace{0.1cm} c$

The function g(x) is continuous at any real number.

Continuity of h(x) = x²

Let’s check the continuity at x = c

Limit = $\lim_{x \to c} h(x) = \lim_{x \to c} (x^2)$

= c²

Function value at x = c, h(c) = c²

As, $\lim_{x \to c} h(x) = h(c) = c^2$

The function h(x) is continuous at any real number.

As, g(x) and h(x) is continuous then g(h(x)) = cos(x²) is also continuous.

问题32.证明由f(x)= |定义的函数cos x |是连续函数。

解决方案：

Let’s take

g(x) = |x|

m(x) = cos x

g(m(x)) = |cos x|

To prove g(m(x)) continuous, g(x) and m(x) should be continuous.

Continuity of g(x) = |x|

As, we know that modulus function works differently.

In |x – 0|, |x| = x when x ≥ 0 and |x| = -x when x < 0

Let’s check the continuity at x = c

When c < 0

Limit = $\lim_{x \to c} g(x) = \lim_{x \to c} (|x|)\\= \lim_{x \to c} (-x)\\ = -c$

Function value at x = c, g(c) = |c| = -c

As, $\lim_{x \to c} g(x) = g(c) = -c$

When c ≥ 0

Limit = $\lim_{x \to c} g(x) = \lim_{x \to c} (|x|)\\= \lim_{x \to c} (x)\\ = c$

Function value at x = c, g(c) = |c| = c

As, $\lim_{x \to c} g(x) = g(c) = c$

The function g(x) is continuous at any real number.

Continuity of m(x) = cos x

Let’s check the continuity at x = c

x = c + h

m(c + h) = cos (c + h)

When x⇢c then h⇢0

cos(A + B) =  cos A cos B – sin A sin B

Limit = $\lim_{h \to 0} m(c+h) = \lim_{h \to 0} (cos(c+h))\\ = \lim_{h \to 0}$(cosc cosh − sinc sinh)

= cosc cos0 − sinc sin0 = cosc

Function value at x = c, m(c) = cos c

As, $\lim_{x \to c} m(x) = m(c) = cos \hspace{0.1cm}c$

The function m(x) is continuous at any real number.

As, g(x) and m(x) is continuous then g(m(x)) = |cos x| is also continuous.

问题33.检查罪孽| x |是连续函数。

解决方案：

Let’s take

g(x) = |x|

m(x) = sin x

m(g(x)) = sin |x|

To prove m(g(x)) continuous, g(x) and m(x) should be continuous.

Continuity of g(x) = |x|

As, we know that modulus function works differently.

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

Let’s check the continuity at x = c

When c < 0

Limit = $\lim_{x \to c} g(x) = \lim_{x \to c} (|x|)\\= \lim_{x \to c} (-x)\\ = -c$

Function value at x = c, g(c) = |c| = -c

As, $\lim_{x \to c} g(x) = g(c) = -c$

When c ≥ 0

Limit = $\lim_{x \to c} g(x) = \lim_{x \to c} (|x|)\\= \lim_{x \to c} (x)\\ = c$

Function value at x = c, g(c) = |c| = c

As, $\lim_{x \to c} g(x) = g(c) = c$

The function g(x) is continuous at any real number.

Continuity of m(x) = sin x

Let’s check the continuity at x = c

x = c + h

m(c + h) = sin (c + h)

When x⇢c then h⇢0

sin(A + B) = sin A cos B + cos A sin B

Limit = $\lim_{h \to 0} m(c+h) = \lim_{h \to 0} (sin(c+h))\\ = \lim_{h \to 0}$(sinc cosh + cosc sinh) 

= sinc cos0 + cos csin0 = sinc

Function value at x = c, m(c) = sin c

As, $\lim_{x \to c} m(x) = m(c) = sin c$

The function m(x) is continuous at any real number.

As, g(x) and m(x) is continuous then m(g(x)) = sin |x| is also continuous.

问题34.找出f(x)= |定义的f的所有不连续点。 x | – | x + 1 |

解决方案：

Let’s take

g(x) = |x|

m(x) = |x + 1|

g(x) – m(x) = | x | – | x + 1 |

To prove g(x) – m(x) continuous, g(x) and m(x) should be continuous.

Continuity of g(x) = |x|

As, we know that modulus function works differently.

In |x – 0|, |x| = x when x≥0 and |x| = -x when x < 0

Let’s check the continuity at x = c

When c < 0

Limit = $\lim_{x \to c} g(x) = \lim_{x \to c} (|x|)\\= \lim_{x \to c} (-x)\\ = -c$

Function value at x = c, g(c) = |c| = -c

As, $\lim_{x \to c} g(x) = g(c) = -c$

When c ≥ 0

Limit = $\lim_{x \to c} g(x) = \lim_{x \to c} (|x|)\\= \lim_{x \to c} (x)\\ = c$

Function value at x = c, g(c) = |c| = c

As, $\lim_{x \to c} g(x) = g(c) = c$

The function g(x) is continuous at any real number.

Continuity of m(x) = |x + 1|

As, we know that modulus function works differently.

In |x + 1|, |x + 1| = x + 1 when x ≥ -1 and |x + 1| = -(x + 1) when x < -1

Let’s check the continuity at x = c

When c < -1

Limit = $\lim_{x \to c} m(x) = \lim_{x \to c} (|x+1|)\\= \lim_{x \to c} -(x+1)$

= -(c + 1)

Function value at x = c, m(c) = |c + 1| = -(c + 1)

As, $\lim_{x \to c} m(x) = m(c) = -(c+1)$

When c ≥ -1

Limit = $\lim_{x \to c} m(x) = \lim_{x \to c} (|x|)\\= \lim_{x \to c} (x+1)$

= c + 1

Function value at x = c, m(c) = |c| = c + 1

As, $\lim_{x \to c} m(x)$ = m(c) = c + 1

The function m(x) is continuous at any real number.

As, g(x) and m(x) is continuous then g(x) – m(x) = |x| – |x + 1| is also continuous.