第 12 类 RD Sharma 解决方案 - 第 9 章连续性 - 练习 9.1 |设置 1

问题1、在原点测试以下函数的连续性：

$f(x)= \begin{cases}\frac{x}{|x|},& x \neq 0 \\1,& x=0\end{cases}$

解决方案：

Given that

$f(x)= \begin{cases}\frac{x}{|x|},& x\neq0 \\1,& x=0\end{cases}$ 

Now, let us consider LHL at x = 0

$\lim_{x\to0^-}f(x) =\lim_{h\to0}f(0-h)$

$=\lim_{h\to0}f(-h)$

$=\lim_{h\to0}\frac{-h}{|-h|}$

$=\lim_{h\to0}\frac{-h}{h} =-1$

Now, let us consider RHL at x = 0

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

$=\lim_{h \to 0}\frac{h}{|h|}=1$

So, LHL ≠ RHL

Therefore, f(x) is discontinuous at origin and the discontinuity is of 1st kind.

问题 2.函数f(x) 定义为$f(x)= \begin{cases}\frac{x^2-x-6}{x-3},& \text{if } x\neq3 \\5,& \text{if }x=3\end{cases}$ .证明 f(x) 在 x = 3 处是连续的。

解决方案：

Given that

 $f(x)= \begin{cases}\frac{x^2-x-6}{x-3},& \text{if } x\neq3 \\5,& \text{if }x=3\end{cases}$

So, here we check the given f(x) is continuous at x = 3,

Now, let us consider LHL at x = 3

$\lim_{x \to 3^-} f(x) =\lim_{h \to 0}f(3-h)$

$=\lim_{h \to 0}\frac{(3-h)^2-(3-h)-6}{(3-h)-3}$

$=\lim_{h \to 0}\frac{h^2-5h}{-h}$

$=\lim_{h\to0}-h+5=5$

Now, let us consider RHL at x = 3

$\lim_{x\to3^+} f(x) =\lim_{h\to0}f(3+h)$

$=\lim_{h\to0}\frac{(3+h)^2-(3+h)-6}{(3+h)-3}$

$=\lim_{h\to0}\frac{h^2+5h}{h}$

$=\lim_{h\to0}h+5=5$

So, f(3) = 5

LHL= RHL = f(3) 

Therefore, f(x) is continuous at x = 3

问题 3.函数f(x) 定义为

$f(x)= \begin{cases}\frac{x^2-9}{x-3},& \text{if } x\neq3 \\6,& \text{if }x=3\end{cases}$

证明 f(x) 在 x = 3 处是连续的。

解决方案：

Given that

$f(x)= \begin{cases}\frac{x^2-9}{x-3},& \text{if } x\neq3 \\6,& \text{if }x=3\end{cases}$

So, here we check the given f(x) is continuous at x = 3,

Now, let us consider LHL at x = 3

$\lim_{x\to3^-}f(x)=\lim_{h\to0}f(3-h)$

$=\lim_{h\to0}\frac{(3-h)^2-9}{(3-h)-3}$

$=\lim_{h\to0}\frac{h^2-6h}{-h}$

$=\lim_{h\to0}-h+6=6$

Now, let us consider RHL at x = 3

$\lim_{x\to3^+}f(x) =\lim_{h\to0}f(3+h)$

$=\lim_{h\to0}\frac{(3+h)^2-9}{(3+h)-3}$

$=\lim_{h\to0}\frac{h^2+6h}{h}$

$=\lim_{h\to0}h+6=6$

So, f(3) = 6

LHL= RHL= f(3)

Therefore, f(x) is continuous at x = 3

问题 4。 $f(x)= \begin{cases}\frac{x^2-1}{x-1},& \text{if } x\neq1 \\2,& \text{if }x=1\end{cases}$

求 f(x) 在 x = 1 处是否连续

解决方案：

Given that

$f(x)= \begin{cases}\frac{x^2-1}{x-1},& \text{if } x\neq1 \\2,& \text{if }x=1\end{cases}$

So, here we check the given f(x) is continuous at x = 1,

Now, let us consider LHL at x = 1

$\lim_{x\to1^-}f(x)=\lim_{h\to0}f(1-h)$

$=\lim_{h\to0}\frac{(1-h)^2-1}{(1-h)-1}$

$=\lim_{h\to0}\frac{1+h)^2-2h-1}{1-h-1}$

$=\lim_{h\to0}\frac{h^2-2h}{-h}$

$=\lim_{h\to0}\frac{h(h-2)}{-h}$

$=\lim_{h\to0}(2-h)=2$

Now, let us consider RHL at x = 1

$\lim_{x\to1^+}f(x)=\lim_{h\to0}f(1+h)$

$=\lim_{h\to0}\frac{(1+h)^2-1}{(1+h)-1}$

$=\lim_{h\to0}\frac{1+h^2+2h-1}{1+h-1}$

$=\lim_{h\to0}\frac{h^2+2h}{h}$

$=\lim_{h\to0}\frac{h(h+2)}{h}$

$=\lim_{h\to0}(2+h)=2$

So, f(1) = 2

LHL= RHL = f(1)

Therefore, f(x) is continuous at x = 1

问题 5. 如果$f(x)= \begin{cases}\frac{sin3x}{x},& \text{when } x\neq0 \\1,& \text{when }x=0\end{cases}$

求 f(x) 在 x = 0 处是否连续。

解决方案：

Given that

$f(x)= \begin{cases}\frac{sin3x}{x},& \text{when } x\neq0 \\1,& \text{when }x=0\end{cases}$

So, here we check the given f(x) is continuous at x = 0,

Now, let us consider LHL at x = 0

$\lim_{x\to0^-}f(x) =\lim_{h\to0}f(0-h)$

$=\lim_{h\to0}\frac{sin3(-h)}{-h}$

$=\lim_{h\to0}-\frac{sin3h}{-h}=3$

Now, let us consider RHL at x = 0

$\lim_{x\to0^+}f(x) =\lim_{h\to0}f(0+h)$

$=\lim_{h\to0}\frac{sin3h}{h}=3$

So, f(0) = 1

LHL = RHL≠ f(0)

Therefore, f(x) is discontinuous at x = 0.

问题 6. 如果$f(x)= \begin{cases}\\e^{\frac{1}{x}},& \text{if } x\neq0 \\1,& \text{if }x=0\end{cases}$

求 f 在 x = 0 处是否连续。

解决方案：

Given that

$f(x)= \begin{cases}\\e^{\frac{1}{x}},& \text{if } x\neq0 \\1,& \text{if }x=0\end{cases}$

So, here we check the given f(x) is continuous at x = 0,

Now, let us consider LHL at x = 0

$\lim_{x\to0^-}f(x) =\lim_{x\to0}f(x)$

$=\lim_{h\to0}f(0-h)$

$=\lim_{h\to0}e^\frac{1}{-h}=e^{-∞}=0$

Now, let us consider RHL at x = 0

$\lim_{x\to0^+}f(x)=\lim_{h\to0}f(0+h)$

$=\lim_{h\to0}e^\frac{1}{h}=e^{-∞}=∞$

So, LHL≠ RHL

Therefore, the f(x) is discontinuous at x = 0. 

问题 7. 让$f(x)= \begin{cases}\frac{1-cosx}{x^2},& \text{when } x\neq0 \\1,& \text{when }x=0\end{cases}$

证明 f(x) 在 x = 0 处不连续。

解决方案：

Given that

$f(x)= \begin{cases}\frac{1-cosx}{x^2},& \text{when } x\neq0 \\1,& \text{when }x=0\end{cases}$

So, here we check the given f(x) is discontinuous at x = 0,

Now, let us consider LHL at x = 0

$\lim_{x\to0^-}f(x) =\lim_{x\to0}f(0-h)$

$=\lim_{h\to0}\frac{1-cos(-h)}{(-h)^2}$

$=\lim_{h\to0}\frac{1-cosh}{h^2}$

$=\lim_{h\to0}\frac{2sin^2(h/2)}{h^2}$

$=\lim_{h\to0}2(\frac{sin(h/2)}{h})^2$

= 2 × 1/4 = 1/2                           

Now, let us consider RHL at x = 0

$\lim_{x\to0^+}f(x) =\lim_{x\to0}f(0+h)$

$=\lim_{h\to0}\frac{1-cosh}{h^2}$

$=\lim_{h\to0}\frac{2sin^2(h/2)}{h^2}$

$=\lim_{h\to0}2(\frac{sin(h/2)}{h})^2$

= 2 × 1/4 = 1/2                           

f(0) = 1

LHL= RHL ≠ f(0)

Therefore, the f(x) is discontinuous at x = 0. 

问题 8. 证明$f(x)= \begin{cases}\frac{x-|x|}{2},& \text{when } x\neq0 \\2,& \text{when }x=0\end{cases}$在 x = 0 处不连续。

解决方案：

Given that

$f(x)= \begin{cases}\frac{x-|x|}{2},& \text{when } x\neq0 \\2,& \text{when }x=0\end{cases}$

So, here we check the given f(x) is discontinuous at x = 0,

Now, let us consider LHL at x = 0

$\lim_{x\to0^-}f(x)=\lim_{h\to0}f(0-h)$

$=\lim_{h\to0}\frac{-h-|h|}{2}$

$=\lim_{h\to0}\frac{-h-h}{2}=0$

Now, let us consider RHL at x = 0

$\lim_{x\to0^+}f(x)=\lim_{h\to0}f(0+h)$

$=\lim_{h\to0}\frac{h-|h|}{2}=0$

f(0) = 2

Thus, LHL= RHL≠ f(0)

Therefore, f(x) is discontinuous at x = 0. 

问题 9. 证明$f(x)= \begin{cases}\frac{|x-a|}{x-a},& \text{when } x\neq a \\1,& \text{when }x=a\end{cases}$在 x = a 处不连续。

解决方案：

Given that

$f(x)= \begin{cases}\frac{|x-a|}{x-a},& \text{when } x\neq a \\1,& \text{when }x=a\end{cases}$

So, here we check the given f(x) is discontinuous at x = a,

Now, let us consider LHL at x = a

$\lim_{x\to a^-}f(x)=\lim_{h\to0}f(a-h)$

$=\lim_{h\to0}\frac{|a-h-a|}{a-h-a}$

$=\lim_{h\to0}\frac{h}{-h}=-1$

Now, let us consider RHL at x = a

$\lim_{x\to a^+}f(x)=\lim_{h\to0}f(a+h)$

$=\lim_{h\to0}\frac{|a+h-a|}{a+h-a}$

$=\lim_{h\to0}\frac{h}{h}=1$

Thus, LHS ≠ RHL

Therefore, the f(x) is discontinuous at x = a.

在指定点讨论以下功能的连续性：

问题 10 (i)。 $f(x)= \begin{cases}|x|cos(\frac{1}{x}),& x\neq0 \\0,& x=0\end{cases}at\space x=0$

解决方案：

Given that

$f(x)= \begin{cases}|x|cos(\frac{1}{x}),& x\neq0 \\0,& x=0\end{cases}at\space x=0$

So, here we check the continuity of the given f(x) at x = 0,

Let us consider LHL,

$\lim_{x\to0^-}f(x)=\lim_{h\to0}f(0-h)$

$=\lim_{h\to0} (|-h|cos(-1/h)$

$=\lim_{h\to0}hcos(-1/h)=0$

Now, let us consider RHL,

$\lim_{x\to0^+}f(x)=\lim_{h\to0}f(0+h)$

$=\lim_{h\to0}|h|cos(1/h)=0$

f(0) = 0

Thus, LHL= RHL= f(0) = 0

Therefore, f(x) is continuous at x = 0.

问题 10 (ii)。 $f(x)= \begin{cases}x^2sin(\frac{1}{x}),& \text{when } x\neq0 \\0,& \text{when }x=0\end{cases}$在 x = 0

解决方案：

Given that

$f(x)= \begin{cases}x^2sin(\frac{1}{x}),& \text{when } x\neq0 \\0,& \text{when }x=0\end{cases}$

So, here we check the continuity of the given f(x) at x = 0,

Let us consider LHL,

$\lim_{x\to0^-}f(x)=\lim_{h\to0}f(0-h)$

$=\lim_{h\to0}(-h)^2sin(-1/h)=0$

Now, let us consider RHL,

$\lim_{x\to0^+}f(x)=\lim_{h\to0}f(0+h)$

$=\lim_{h\to0}h^2sin(1/h)=0$

f(0) = 0

Thus, LHL= RHL = f(0) = 0

Therefore, f(x) is continuous at x = 0.

问题 10 (iii)。 $f(x)= \begin{cases}(x-a)sin(\frac{1}{x-a}),& \text{when } x\neq a \\0,& \text{when }x=a\end{cases}$在 x = a

解决方案：

Given that

$f(x)= \begin{cases}(x-a)sin(\frac{1}{x-a}),& \text{when } x\neq a \\0,& \text{when }x=a\end{cases}$

So, here we check the continuity of the given f(x) at x = a,

Let us consider LHL,

$\lim_{x\to a^-}f(x)=\lim_{h\to0}f(a-h)$

$=\lim_{h\to0}(a-h-a)sin(\frac{1}{a-h-a})$

$=\lim_{h\to0}-hsin(1/h)=0$

Now, let us consider RHL,

$\lim_{x\to a^+}f(x)=\lim_{h\to0}f(a+h)$

$=\lim_{h\to0}(a+h-a)sin(\frac{1}{a+h-a})$

$=\lim_{h\to0}hsin(1/h)=0$

f(a) = 0

Thus, LHL= RHL= f(a) = 0

Therefore, f(x) is continuous at x = 0.

问题 10 (iv)。 $f(x)= \begin{cases}\frac{(e^x-1)}{log(1+2x)},& \text{when } x\neq0 \\7,& \text{when }x=0\end{cases}$在 x = 0

解决方案：

Given that

$f(x)= \begin{cases}\frac{(e^x-1)}{log(1+2x)},& \text{when } x\neq0 \\7,& \text{when }x=0\end{cases}$

So, here we check the continuity of the given f(x) at x = 0,

$\lim_{x\to0}f(x)=\lim_{x\to0}\frac{e^x-1}{log(1+2x)}$

$=\lim_{x\to0}\frac{e^x-1}{\frac{2xlog(1+2x)}{2x}}$

$=(1/2)×\lim_{x\to0}\frac{(e^x-1)/x}{\frac{log(1+2x)}{2x}}$

$=(1/2)×\frac{\lim_{x\to0}(e^x-1)/x}{\lim_{x\to0}\frac{log(1+2x)}{2x}}$ 

= 1/2 × 1/1 = 1/2                        

And, 

f(0) = 7

$\lim_{x\to0}f(x)$ ≠ f(0)

Therefore, f(x) is discontinuous at x = 0.

问题 10 (v)。 $f(x)= \begin{cases}\frac{(1-x^n)}{(1-x)},& \text{when } x\neq1 \\n-1,& \text{when }x=1\end{cases}$ n ∈ N 在 x = 1

解决方案：

Given that

$f(x)= \begin{cases}\frac{(1-x^n)}{(1-x)},& \text{when } x\neq1 \\n-1,& \text{when }x=1\end{cases}$

So, here we check the continuity of the given f(x) at x = 1,

Let us consider LHL,

$\lim_{x\to1^-}f(x) =\lim_{h\to0}f(1-h)$

$=\lim_{h\to0}\frac{1-(1-h)^n}{1-(1-h)}$

$=\lim_{h\to0}\frac{1-[1-nh+\frac{n(n-1)}{2!}h^2+...]}{h}$

$=\lim_{h\to0}n-\frac{n(n-1)h}{2!}+...=n$

Now, let us consider RHL,

$\lim_{x\to1^+}f(x) =\lim_{h\to0}f(1+h)$

$=\lim_{h\to0}\frac{1-(1+h)^n}{1-(1+h)}$ 

$=\lim_{h\to0}\frac{1-[1+nh+\frac{n(n-1)}{2!}h^2+...]}{-h}$

$=\lim_{h\to0}n+\frac{n(n-1)h}{2!}+...=n$

f(1) = n – 1

Thus, LHL = RHL ≠ f(1)

Therefore, f(x) is discontinuous at x = 1.

问题 10 (vi)。 $f(x)= \begin{cases}\frac{|x^2-1|}{x-1},& \text{when } x\neq1 \\2,& \text{when }x=1\end{cases}$在 x = 1

解决方案：

Given that

$f(x)= \begin{cases}\frac{|x^2-1|}{x-1},& \text{when } x\neq1 \\2,& \text{when }x=1\end{cases}$

So, here we check the continuity of the given f(x) at x = 1,

Let us consider LHL,

$\lim_{x\to1^-}f(x) =\lim_{h\to0}f(1-h)$

$=\lim_{h\to0}\frac{|(1-h)^2-1|}{(1-h)-1}$

$=\lim_{h\to0}\frac{|h^2-2h|}{-h}$

$=\lim_{h\to0}(h-2)=-2$

Now, let us consider RHL,

$\lim_{x\to1^+}f(x) =\lim_{h\to0}f(1+h)$

$=\lim_{h\to0}\frac{|(1+h)^2-1|}{(1+h)-1}$ 

$=\lim_{h\to0}\frac{h^2+2h}{h}=2$

f(1) = 2

LHL= RHL = f(1) = 2

Therefore, f(x) is discontinuous at x = 1.

问题 10 (vii)。 $f(x)= \begin{cases}\frac{2|x|+x^2}{x},& \text{when } x\neq0 \\0,& \text{when }x=0\end{cases}$在 x = 0

解决方案：

Given that

$f(x)= \begin{cases}\frac{2|x|+x^2}{x},& \text{when } x\neq0 \\0,& \text{when }x=0\end{cases}$

So, here we check the continuity of the given f(x) at x = 0,

Let us consider LHL,

$\lim_{x\to0^-}f(x) =\lim_{h\to0}f(0-h)$

$=\lim_{h\to0}\frac{2(|-h|)+(-h)^2}{-h}$

$=\lim_{h\to0}\frac{2h+h^2}{-h}=-2$

Let us consider RHL,

$\lim_{x\to0^+}f(x) =\lim_{h\to0}f(0+h)$

$=\lim_{h\to0}\frac{2×|h|+h^2}{h}=2$

Thus, LHL ≠ RHL

Therefore, f(x) is discontinuous at x = 0.

问题 10 (viii)。 $f(x)= \begin{cases}|x-a|sin\frac{1}{(x-a)},& \text{when } x\neq a \\0,& \text{when }x=a\end{cases}$在 x = a

解决方案：

Given that, 

$f(x)= \begin{cases}|x-a|sin\frac{1}{(x-a)},& \text{when } x\neq a \\0,& \text{when }x=a\end{cases}$

f(x) = (x – a)sin{1/(x – a)}, x > 0

= (x – a)sin{1/(x – a)}, x < 0

= 0, x = a 

Let us consider LHL,

$\lim_{x\to a^-}f(x)=-(a+a)sin(\frac{1}{-a+a})=0$

Now, let us consider RHL,

$\lim_{x\to a^+}f(x)=(a-a)sin(\frac{1}{a-a})=0$

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

Therefore, f(x) is continuous at x = a.

问题 11. 证明$f(x)=\begin{cases}1+x^2,& \text{if }0\leq x\leq1 \\2-x,& \text{if }x>1\end{cases}$在 x = 1 处不连续。

解决方案：

Given that, 

$f(x)=\begin{cases}1+x^2,& \text{if }0\leq x\leq1 \\2-x,& \text{if }x>1\end{cases}$

So, here we check the given f(x) is discontinuous at x = 1,

Let us consider LHL,

$\lim_{x\to1^-}f(x) =\lim_{h\to0}f(1-h)$

$=\lim_{h\to0}1+(1-h)^2$

$=\lim_{h\to0}1+1-2h+h^2=2$

Now, let us consider RHL,

$\lim_{x\to1^+}f(x) =\lim_{h\to0}f(1+h)$

$=\lim_{h\to0}2-(1+h)=1$

LHL ≠ RHL

Therefore, f(x) is discontinuous at x = 1.

问题 12. 证明$f(x)= \begin{cases}\frac{sin3x}{tan2x},& \text{if }x<0 \\(\frac{3}{2})&\text{if }x=0\\\frac{log(1+3x)}{e^{2x}-1},&\text{if }x>0\end{cases}$在 x = 0 处是连续的

解决方案：

Given that,

$f(x)= \begin{cases}\frac{sin3x}{tan2x},& \text{if }x<0 \\(\frac{3}{2})&\text{if }x=0\\\frac{log(1+3x)}{e^{2x}-1},&\text{if }x>0\end{cases}$

So, here we check the given f(x) is continuous at x = 0,

Let us consider LHL,

$\lim_{x\to0^-}f(x) =\lim_{h\to0}f(0-h)$

$=\lim_{h\to0}\frac{sin(3(-h))}{tan(2(-h))}$

$=\lim_{h\to0}\frac{-sin3h}{-tan2h}$

$=\lim_{h\to0}\frac{\frac{sin3h}{3h}3h}{\frac{tan2h}{2h}2h}=3/2$

Let us consider RHL,

$\lim_{x\to0^+}f(x) =\lim_{h\to0}f(0+h)$

$=\lim_{h\to0}\frac{log(1+3h)}{e^{2h}-1}$

$=\lim_{h\to0}\frac{\frac{log(1+3h)}{3h}3h}{\frac{e^{2h}-1}{2h}2h}=3/2$

f(0) = 3/2

Thus, LHL = RHL = f(0) = 3/2

Therefore, f(x) is continuous at x = 0.

问题 13. 求函数f 定义为 'a' 的值

$f(x)=\begin{cases}asin\frac{π}{2}(x+1),& \text{if }x\leq0 \\\frac{tanx-sinx}{x^3},& \text{if }x>0\end{cases}$在 x = 0 处是连续的。

解决方案：

Given that,

$f(x)=\begin{cases}asin\frac{π}{2}(x+1),& \text{if }x\leq0 \\\frac{tanx-sinx}{x^3},& \text{if }x>0\end{cases}$

Let us consider LHL,

$\lim_{x\to0^-}f(x) =\lim_{h\to0}f(0-h)$

$=\lim_{h\to0}f(-h)$

$=\lim_{h\to0}asin(\frac{π}{2})(-h+1)$

$=\lim_{h\to0}asin\frac{π}{2}=a$

Now, let us consider RHL,

$\lim_{x\to0^+}f(x) =\lim_{h\to0}f(0+h)=\lim_{h\to0}f(h)=\lim_{h\to0}\frac{tanh-sinh}{h^3}$

⇒ $\lim_{h\to0^+}f(x)=\lim_{h\to0}\frac{\frac{sinh}{cosh}-sinh}{h^3}$

$=\lim_{h\to0}\frac{\frac{sinh}{cosh}(1-cosh)}{h^3}$

$=\lim_{h\to0}\frac{(1-cosh)tanh}{h^3}$

$=\lim_{h\to0}\frac{2sin^2(\frac{h}{2})tanh}{4(\frac{h^2}{4})×h}$

$=(2/4)\lim_{h\to0}\frac{sin^2(\frac{h}{2})tanh}{(\frac{h^2}{4})×h}$

$=(1/2)\lim_{h\to0}(\frac{sin(\frac{h}{2})}{\frac{h}{2}})^2\lim_{h\to0}\frac{tanh}{h}$

= (1/2) × 1 × 1

⇒ $\lim_{x\to0^+}f(x)=1/2$

If f(x) is continuous at x = 0, then

$\lim_{x\to0^-}f(x)=\lim_{x\to0^+}f(x)$

⇒ a = 1/2

问题 14. 检查函数的连续性

$f(x)=\begin{cases}3x-2,& \text{if }x\leq 0 \\x+1,& \text{if }x>0\end{cases}$在 x = 0

还要画出这个函数的图形。

解决方案：

Given that, 

$f(x)=\begin{cases}3x-2,& \text{if }x\leq 0 \\x+1,& \text{if }x>0\end{cases}$

So, here we check the continuity of the given f(x) at x = 0,

Let us consider LHL,

$\lim_{x\to0^-}f(x) =\lim_{h\to0}f(0-h)$

$=\lim_{h\to0}3(-h)-2$

$=\lim_{h\to0}-3h-2=-2$

Now, let us consider RHL,

$\lim_{x\to0^+}f(x) =\lim_{h\to0}f(0+h)$

$=\lim_{h\to0}h+1=1$

LhL ≠ RHL

So, the f(x) is discontinuous.

问题 15. 讨论函数的连续性

$f(x)=\begin{cases}x,& \text{if }x>0 \\1,& \text{if }x=0 \\-x,& \text{if }x<0\end{cases}$在点 x = 0。

解决方案：

Given that, 

$f(x)=\begin{cases}x,& \text{if }x>0 \\1,& \text{if }x=0 \\-x,& \text{if }x<0\end{cases}$

So, here we check the continuity of the given f(x) at x = 0,

Let us consider LHL,

$\lim_{x\to0^-}f(x) =\lim_{h\to0}f(0-h)$

$=\lim_{h\to0}(-h)=0$

Now, let us consider RHL,

$\lim_{x\to0^+}f(x) =\lim_{h\to0}f(0+h)$

$=\lim_{h\to0}h=0$

f(0) = 1

LHL = RHL ≠ f(0)

Hence, the f(x) is discontinuous at x = 0.