Limit And Continuity

1. Define limit of a function at a point.
Ans: The limit of a function at a point k in its domain (if it exists) is the value that the function approaches as its argument approaches k.

2. Write the notational representation of 'x approches to 5.
Ans: Here, 
    The   notational representation of 'x approches to 5 is  x ⟶ 5.

3. What is meaning of   x ⟶ a ?
Ans: Meaning of  x ⟶ a is   'x approches to a' or 'x tends to a'.

4. What do you understand by indeterminate form?
Ans: In limits involving an algebraic combination of functions in an independent variable may often be evaluated by replacing these functions by their limits;
if the expression obtained after this substitution does not give enough information to determine the original limit, it is said to take on an indeterminate form.
Which is also called ∞ .

5. Evaluate the following limit:
        ⁡limx→4(10x2) 
Solution: Here,
   ⁡limx→4(10x2)
=  10 ⨯ 4²
= 160.

6. 

Calculate:
⁡limx→1(5 - x2).

Solution: Here, 

  ⁡limx→1(5 - x2)

= 5 - 1²
= 5 - 1
= 4.

7. 

Evaluate the following:
       limx→-2(10 - x3).

Solution: Here,

       ⁡limx→-2(10 - x3)
=  10 - (-2)³
= 10 - (-8)
= 10 + 8
= 18.

8. 

Calculate:
            
  ⁡limx→2(x2 - 4)

Solution: Here, 

        ⁡limx→2(x2- 4)
= 2² - 4
= 4 - 4
= 0.

9. 

Evaluate the following limit:
              limx→-1(40 - 50x4).

Solution: Here, 

         
⁡limx→-1(40 - 50x4)
           = 40 - 50(-1)⁴
           = 40 - 50 
           = - 10.
 

10. 

Evaluate :  ⁡limx→a(a² - x2).

Solution: Here,

   ⁡limx→a(a² - x2)
= a² - a²
= 0.

11. 

Calcultae the value of :  
⁡limx→-a(10x2- 5a²).

Solution: Here,

      ⁡limx→- a (10x2-5a²)
= 10(-a)² - 5a²
= 10a² - 5a²
= 5a².

12. 

Evaluate the following limit:
⁡limx→- 3 f(x)   : f(x) = (x³- 10x2- x +5). 

Solution: Here, ⁡

limx→- 3 f(x)     : f(x) = x³- 10x2- x + 5
          = 
⁡limx→-3 (x³ - 10x2- x +5)
          = (-3)³ - 10(-3)² -(-3) + 5
          = - 27 - 30 + 3 + 5
          = -49.

13. 

Evaluate : 
⁡limx→5(x³ -10x2).

Solution: Here,

        ⁡limx→5(x³ - 10x2).
   =    (5)³ - 10(5)²
   = 125 - 250
   =  - 125.

 

14. Evaluate the following:
⁡  lim     x→-1x3 + 1  x + 1

Solution: 

Here,  lim     x→-1x3 + 1  x + 1
      ⁡

= 

15. 

Calculate: ⁡limx→4x² - 4xx - 4

16. 

Evaluate :  ⁡limx→-1x3 - 1x + 1.

17. 

Calculate:    limx→3⁡ x² - 9x² - 3x

18. 

Find the value of f(x) =    x2 + 1x³ + 1   at  x = 2.

Solution: Here,f(x) =   x2 + 1x³ + 1
At x = 2,
f(2) =   2² + 12³ + 1 
       =   59 .

19.  Evaluate : 

limx→ k ⁡x3 + k³x + k .

Solution: Here, ⁡limx→k x3 + k³x + k
   =     k³+ k³ k + k 
    =      2k³2k
    =   k².

20. 

Evaluate , ⁡limx→-k x3 + k³x + k

Solution: 

21. 

Examine whether the following functions are defined or not at the point  x = 2. If defined , find the value.
f(x) = 

1x - 2 + 6x8 - x³ 

Solution: Here, f(x) = 

1x - 2 + 6x8 - x³ 
At x = 2,
  f(2) = 1 2 - 2 + 6(2) 8 - 2³
       = 10 + 120
      = ∞ + ∞
      = ∞, which is infinite number.
Thus, f(x) is not defined at x  = 2.

22. 

Examine whether the following functions are defined or not at the point  x = 5. If defined , find the value.
f(x) = 

(x + 1)⁴ - 1x

Solution: Here, f(x) = 

(x + 1)⁴ - 1x
At x = 5,
  f(5) = (5 + 1)⁴ - 15
        = 6⁴ - 15
       = 12955
       = 259 , which is finite number.
Thus f(x) is defined at x = 5 and the value of function at x = 5 is 259.

23. 

Find : ⁡limx→35x² by using table.

Solution: Let f(x) = 5x²  & a = 3.

We choose the values of x close to 3.
 

x	f(x) = 5x2
2.9	5 ⨯ 2.92 = 42.05
2.99	5 ⨯ 2.992 = 44.701
2.999	5 ⨯ 2.9992 = 44.97
..........	......................  ?
3.0001	5 ⨯ 3.00012 = 45.003
3.001	5 ⨯ 3.0012 = 45.03
3.01	5 ⨯ 3.012 = 45.301

From the table, we infer that;
As   x ⟶ 3, f(x) ⟶45.
Therefore, 
          ⁡limx→35x² = 45.
Thus, the required limit is 45.

24. 

Find: 
⁡lim  x→2x² - 4  x - 2 ; by using table.

Solution: Here, 

    Let  f(x) = ⁡limx→2x² - 4x - 2    and a = 2.
x = 2 makes the given function indeterminate( 0 / 0) form.
We cjoose the values of x close to 2.
 

x	f(x) = (x2 - 4) / (x - 2)
1.99	3.99
1.999	3.999
1.9999	3.9999
.............(2)	..............?      (4)
2.0001	4.0001
2.001	4.001
2.01	4.01

As  x ⟶ 2, then f(x) ⟶ 4.
Thus,  ⁡limx→2x² - 4x - 2 = 4.

25. 

Fill the table given below and also find the limiting value of function f(y) = y - 1y² + 1  at y⟶2.
 

y	1.99	1.999	1.9999	2.00001	2.0001	2.001
f(y)	.....	.......	.......	.......	......	.......

Solution: Here,  

f(y) = y - 1y² + 1   
Now, completing the table;

y	1.99	1.999	1.9999	2.00001	2.0001	2.001
f(y)	0.1996	0.19994	0.199999	0.200004	0.20004.	0.2004

From the table, 
Limiting value of f(y) at y = 2 is 0.2.

26. 

Given that: f(x) =  x² - 4x - 2 
a) Does f(2) represent a real number?
b) What are the values of f(x) at x = 1.9, 1.09 & 1.009?

Solution: Here,

     f(x)  = x² - 4x- 2 
When, x = 2,
    f(2) = 2² - 42 - 2
           = 4 - 4 2 - 2
           = 0 0  , (Indeterminate form) which is not a real number.

b) When, x = 1.9, 
     f(1.9)  =  3.9.
When,  x = 1.09,
    f(1.09) = 3.09.

When, x = 1.009

    f(x) = 3.009.