Question 1 to 12:

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Q1. Prove that the function

$$f(x)=5x-3$$

is continuous at $x=0$, $x=-3$ and $x=5$.

Solution

The function $f(x)=5x-3$ is a polynomial function.
We know from theory that every polynomial function is continuous at every real number.  

To verify at specific points:

At $x=0$

$$f(0)=5(0)-3=-3$$
$$\underset{x\to 0}{\lim }f(x)=\underset{x\to 0}{\lim }(5x-3)=5(0)-3=-3$$

Since

$$\underset{x\to 0}{\lim }f(x)=f(0)\Rightarrow f(x)\text{ is continuous at }x=0$$

At $x=-3$

$$f(-3)=5(-3)-3=-18$$
$$\underset{x\to -3}{\lim }(5x-3)=5(-3)-3=-18$$

Thus

$$\underset{x\to -3}{\lim }f(x)=f(-3)$$

At $x=5$

$$f(5)=5(5)-3=22$$
$$\underset{x\to 5}{\lim }(5x-3)=22$$

Conclusion

$$f(x)\text{ is continuous at }x=0,-3,5$$

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Q2. Examine continuity of

$$f(x)=2x^2-1\text{ at }x=3$$

Solution

The function is defined at $x=3$:

$$f(3)=2(3{)}^2-1=2(9)-1=17$$

Now find the limit:

$$\underset{x\to 3}{\lim }f(x)=\underset{x\to 3}{\lim }(2x^2-1)=2(3{)}^2-1=17$$

Since

$$\underset{x\to 3}{\lim }f(x)=f(3)$$

Therefore, the function is continuous at $x=3$.

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Q3. Examine the following functions for continuity

(a) $f(x)=x-5$

Polynomial ⇒ continuous everywhere.

$$\underset{x\to c}{\lim }(x-5)=c-5=f(c)$$

Continuous for all real x

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(b) $f(x)=\frac{1}{x-5},x\mathrm{\ne }5$

The function is undefined at $x=5$, so its domain excludes 5.
Every rational function is continuous in its domain (Example 16)

Continuous for all $x\mathrm{\ne }5$
Discontinuous at $x=5$

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(c)

$$f(x)=\frac{x^2-25}{x+5},x\mathrm{\ne }-5$$

Factor numerator:

$$x^2-25=(x-5)(x+5)$$

Thus,

$$f(x)=x-5,x\mathrm{\ne }-5$$

This becomes a linear function, but $x=-5$ is excluded from the domain.

Continuous for $x\mathrm{\ne }-5$
Discontinuous at $x=-5$

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(d) $f(x)=\mathrm{\mid }x-5\mathrm{\mid }$

Modulus function is continuous everywhere (Example 7)

 Continuous for all real x

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Q4. Prove that

$$f(x)=x^n\text{ is continuous at }x=n$$

(where $n$ is a positive integer)

Solution

$$f(n)=n^n$$
$$\underset{x\to n}{\lim }{x}^n=n^n$$

Thus

$$\underset{x\to n}{\lim }f(x)=f(n)$$

So the function is continuous at $x=n$

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Q5. Function

$$f(x)=\{\begin{array}{ll}x,& x\le 1\\ 5,& x>1\end{array}$$

Check continuity at $x=0$, $x=1$, $x=2$

At $x=0$

$$f(0)=0$$

Nearby values belong to $x\le 1$:

$$\underset{x\to 0}{\lim }f(x)=\underset{x\to 0}{\lim }x=0=f(0)$$

Continuous at 0

At $x=1$

Left-hand limit:

$$\underset{x\to 1^{-}}{\lim }f(x)=1$$

Right-hand limit:

$$\underset{x\to 1^{+}}{\lim }f(x)=5$$

$$f(1)=1$$

Since

$$LHL=1,RHL=5,LHL\mathrm{\ne }RHL$$

Discontinuous at $x=1$

At $x=2$

$$f(2)=5,\underset{x\to 2}{\lim }f(x)=5$$

Continuous at $x=2$

Point of discontinuity

$$x=1$$

Find all points of discontinuity of f, where f is defined by

Q6. The function is:

$$f(x)=\{\begin{array}{ll}2x+3,& x\le 2\\ 2x-3,& x>2\end{array}$$

We must examine continuity at $x=2$.

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Step 1: Check if the function is defined at $x=2$

Since $x=2$ satisfies $x\le 2$, we use the first expression:

$$f(2)=2(2)+3=4+3=7$$

So,$$f(2)=7$$

Thus, the function is defined at $x=2$.

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Step 2: Left-Hand Limit (LHL) at $x=2$

$$\underset{x\to 2^{-}}{\lim }f(x)=\underset{x\to 2^{-}}{\lim }(2x+3)$$
$$=2(2)+3=4+3=7$$

So,

$$LHL=7$$

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Step 3: Right-Hand Limit (RHL) at $x=2$

$$\underset{x\to 2^{+}}{\lim }f(x)=\underset{x\to 2^{+}}{\lim }(2x-3)$$
$$=2(2)-3=4-3=1$$

So,

$$RHL=1$$

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Step 4: Compare limits and value at the point

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Condition for continuity

$$f(x)\text{ is continuous at }x=a\text{ if }\underset{x\to a}{\lim }f(x)=f(a)$$

Since

$$LHL\mathrm{\ne }RHL\Rightarrow \text{Limit does not exist at }x=2$$

So, the function cannot be continuous at that point.

FINAL ANSWER

$$\boxed{{\displaystyle f(x)\text{ is discontinuous at }x=2.}}$$

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Question 7 

The function is defined as:

$$f(x)=\{\begin{array}{ll}\mathrm{\mid }x\mathrm{\mid }+3,& x\le -3\\ -2x,& -3<x<3\\ 6x+2,& x\ge 3\end{array}$$

We must check continuity at the points where the definition changes, i.e.,

$$x=-3\text{and}x=3$$

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Check Continuity at $x=-3$

Step 1: Find $f(-3)$

Since $-3\le -3$, use $\mathrm{\mid }x\mathrm{\mid }+3$:

$$f(-3)=\mathrm{\mid }-3\mathrm{\mid }+3=3+3=6$$

Step 2: Left-hand limit (LHL) at $x=-3$

For $x<-3$, expression is $\mathrm{\mid }x\mathrm{\mid }+3$

$$\underset{x\to -3^{-}}{\lim }(\mathrm{\mid }x\mathrm{\mid }+3)=\mathrm{\mid }-3\mathrm{\mid }+3=3+3=6$$

Step 3: Right-hand limit (RHL) at $x=-3$

For $-3<x<3$, expression is $-2x$

$$\underset{x\to -3^{+}}{\lim }(-2x)=-2(-3)=6$$

Compare values

$$\text{Since }LHL=RHL=f(-3),\boxed{{\displaystyle \text{f(x) is continuous at }x=-3}}$$

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Check Continuity at $x=3$

Step 1: Find $f(3)$

Since $x\ge 3$, use $6x+2$:

$$f(3)=6(3)+2=18+2=20$$

Step 2: Left-hand limit (LHL)

For $x<3$ and $x>-3$, use $-2x$

$$\underset{x\to 3^{-}}{\lim }(-2x)=-2(3)=-6$$

Step 3: Right-hand limit (RHL)

$$\underset{x\to 3^{+}}{\lim }(6x+2)=6(3)+2=20$$

Compare values

Since

$$LHL\mathrm{\ne }RHL$$
$$\boxed{{\displaystyle \text{f(x) is NOT continuous at }x=3}}$$

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Question 8 

The function is defined as:

$$f(x)=\{\begin{array}{ll}\frac{\mathrm{\mid }x\mathrm{\mid }}{x},& x\mathrm{\ne }0\\ 0,& x=0\end{array}$$

We need to check continuity at $x=0$ because that is the only point where the definition changes.

Step 1: Find $f(0)$

Given directly:

$$f(0)=0$$

Step 2: Find Left-Hand Limit (LHL) as $x\to 0^{-}$

When $x<0$, $\mathrm{\mid }x\mathrm{\mid }=-x$

So,

$$f(x)=\frac{\mathrm{\mid }x\mathrm{\mid }}{x}=\frac{-x}{x}=-1$$

Thus,

$$\underset{x\to 0^{-}}{\lim }\frac{\mathrm{\mid }x\mathrm{\mid }}{x}=-1$$

Step 3: Find Right-Hand Limit (RHL) as $x\to 0^{+}$

When $x>0$, $\mathrm{\mid }x\mathrm{\mid }=x$

So,

$$f(x)=\frac{\mathrm{\mid }x\mathrm{\mid }}{x}=\frac{x}{x}=1$$

Thus,

$$\underset{x\to 0^{+}}{\lim }\frac{\mathrm{\mid }x\mathrm{\mid }}{x}=1$$

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Step 4: Compare values

Condition for Continuity:

$$\text{Function is continuous at }x=a\text{ if }LHL=RHL=f(a)$$

But here:

$$LHL=-1\mathrm{\ne }RHL=1$$

Therefore:

$$\underset{x\to 0}{\lim }f(x)\text{ does not exist}$$So,

$$\boxed{{\displaystyle \text{f(x) is NOT continuous at }x=0}}$$

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Question 9

The given function is:

$$f(x)=\{\begin{array}{ll}\frac{x}{\mathrm{\mid }x\mathrm{\mid }},& x<0\\ -1,& x\ge 0\end{array}$$

We must examine continuity at $x=0$, because that is the point where the rule changes.

Step 1: Find $f(0)$

Since $0\ge 0$, use the second part of the definition:

$$f(0)=-1$$

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Step 2: Find Left-Hand Limit (LHL) as $x\to 0^{-}$

For $x<0$, the expression is:

$$f(x)=\frac{x}{\mathrm{\mid }x\mathrm{\mid }}$$

When $x<0$, $\mathrm{\mid }x\mathrm{\mid }=-x$, so:

$$\frac{x}{\mathrm{\mid }x\mathrm{\mid }}=\frac{x}{-x}=-1$$

Thus:

$$\underset{x\to 0^{-}}{\lim }f(x)=-1$$

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Step 3: Find Right-Hand Limit (RHL) as $x\to 0^{+}$

For $x>0$, use second part of definition (since all $x\ge 0$ map to $f(x)=-1$:

$$f(x)=-1$$

Thus:

$$\underset{x\to 0^{+}}{\lim }f(x)=-1$$

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Step 4: Compare values

So,

$$LHL=RHL=f(0)$$

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Final Conclusion

$$\boxed{{\displaystyle \text{The function is continuous at }x=0}}$$

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FINAL ANSWER

$$\boxed{{\displaystyle f(x)\text{ is continuous everywhere, including at }x=0.}}$$

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Question 10

The function is:$$f(x)=\{\begin{array}{ll}x+1,& x\ge 1\\ x^2+1,& x<1\end{array}$$

We need to examine continuity at $x=1$, because that is the point where the definition switches.

Step 1: Check whether $f(x)$ is defined at $x=1$

Since $1\ge 1$, use first expression:

$$f(1)=1+1=2$$

So, the function is defined at $x=1$, and $\boxed{{\displaystyle f(1)=2}}$

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Step 2: Find the Left-Hand Limit (LHL) as $x\to 1^{-}$

For values less than 1, use $x^2+1$:

$$\underset{x\to 1^{-}}{\lim }f(x)=\underset{x\to 1^{-}}{\lim }(x^2+1)=1^2+1=2$$

So,

$$\boxed{{\displaystyle LHL=2}}$$

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Step 3: Find the Right-Hand Limit (RHL) as $x\to 1^{+}$

For values greater than or equal to 1, use $x+1$:

$$\underset{x\to 1^{+}}{\lim }f(x)=\underset{x\to 1^{+}}{\lim }(x+1)=1+1=2$$

So,

$$\boxed{{\displaystyle RHL=2}}$$

Step 4: Compare LHL, RHL and $f(1)$

Since

$$LHL=RHL=f(1)$$

FINAL CONCLUSION

$$\boxed{{\displaystyle \text{The function is continuous at }x=1.}}$$

Final Answer

$$\boxed{{\displaystyle f(x)\text{ is continuous at }x=1}}$$

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Question 11

The given function is:$$f(x)=\{\begin{array}{ll}{x}^3-3,& x\le 2\\ x^2+1,& x>2\end{array}$$

We must check continuity at $x=2$, because that is where the definition changes.

Step 1: Check if the function is defined at $x=2$

Since $2\le 2$, we use the first expression:

$$f(2)=2^3-3=8-3=5$$

So,

$$\boxed{{\displaystyle f(2)=5}}$$

The function is defined at $x=2$.

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Step 2: Left-Hand Limit (LHL) as $x\to 2^{-}$

For $x<2$, use $x^3-3$:

$$\underset{x\to 2^{-}}{\lim }f(x)=\underset{x\to 2^{-}}{\lim }(x^3-3)=2^3-3=8-3=5$$

So,

$$\boxed{{\displaystyle LHL=5}}$$

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Step 3: Right-Hand Limit (RHL) as $x\to 2^{+}$

For $x>2$, use $x^2+1$:

$$\underset{x\to 2^{+}}{\lim }f(x)=\underset{x\to 2^{+}}{\lim }(x^2+1)=2^2+1=4+1=5$$

So,$$\boxed{{\displaystyle RHL=5}}$$

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Step 4: Compare Limits and Function Value

Condition for continuity:

$$f(x)\text{ is continuous at }x=a\text{ if }LHL=RHL=f(a)$$

Here:

$$LHL=RHL=f(2)$$

FINAL CONCLUSION

$$\boxed{{\displaystyle f(x)\text{ is continuous at }x=2}}$$

FINAL ANSWER

$$\boxed{{\displaystyle \text{The function }f(x)\text{ is continuous at }x=2.}}$$

Question 12 

The function is:

$$f(x)=\{\begin{array}{ll}{x}^{10}-1,& x\le 1\\ x^2,& x>1\end{array}$$

We need to examine continuity at $x=1$ (the point where the definition changes).

Step 1: Check whether the function is defined at $x=1$

Since $1\le 1$, we use the first expression $x^{10}-1$:

$$f(1)=1^{10}-1=1-1=0$$

So,$$\boxed{{\displaystyle f(1)=0}}$$

Step 2: Left-Hand Limit (LHL) at $x=1$

For values $x<1$, use $x^{10}-1$

$$\underset{x\to 1^{-}}{\lim }f(x)=\underset{x\to 1^{-}}{\lim }(x^{10}-1)=1^{10}-1=0$$

So:

$$\boxed{{\displaystyle LHL=0}}$$

Step 3: Right-Hand Limit (RHL) at $x=1$

For values $x>1$, use $x^2$

$$\underset{x\to 1^{+}}{\lim }f(x)=\underset{x\to 1^{+}}{\lim }(x^2)=1^2=1$$

So:

$$\boxed{{\displaystyle RHL=1}}$$

Step 4: Compare values

Condition for continuity

$$\text{Function is continuous at }x=a\text{ if }LHL=RHL=f(a)$$

But here:

$$LHL=0\mathrm{\ne }1=RHL$$

So the limit does not exist at $x=1$

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FINAL CONCLUSION

$$\boxed{{\displaystyle \text{f(x) is NOT continuous at }x=1}}$$
$$\boxed{{\displaystyle \text{The function is discontinuous at }x=1}}$$