# CPT Chapter Limits and Continuity

*With Anand.Duration:4hrs25min*

Limits and Continuity | Duration (min:sec) |

Limits | |

Overview | 07:49 |

What is Limit | 18:21 |

Limit with graph | 03:35 |

Basics of Limit | 24:27 |

Avoid Indeterminant1 | 24:41 |

Avoid Indeterminant2 | 22:02 |

How to solve infinity problem | 02:14 |

Infinity Problem | 21:03 |

Limit Formulae | 06:19 |

Formulae Example1 | 16:42 |

Formulae Example2 | 29:09 |

Odd & Even Function | 07:07 |

Continuity | |

Continuity Concept | 14:25 |

Function Discontinuous | 03:54 |

Checking Limit Exist Concept | 03:57 |

Example Checking Limit Exist | 12:58 |

Check Continuity-Concept | 02:40 |

Check Countinity-Example1 | 17:28 |

Check Countinity-Example2 | 23:18 |

Conclusion | 02:37 |

Total | 04:24:46 |

What are Limits?

Basic of Limit

Avoid Indeterminant

Infinity Problem

Formula based Problem

Condition for discontinuous

Check Limit Exist or Not

Check Continuous or Not

Avoid Indeterminant-

lim (4x4+5x3+7x2+6x)/(5x5+7x2+x) is equal to

x →0

a) 6 b) 5 c) -6 d) none of these

lim(x2-5x+6)(x2-3x+2)/(x3-3x2+4) is equal to

x →2

a)1/3 b)3 c) -⅓ ) none of these

lim (4-x2)/3-√(x2+5) is equal to

x →2

a) 6 b) 1/6 c) –6 d) none of these

lim (x2-1)/(√3x+1-√5x-1) is evaluated to be

x →1

a) 4 b) ¼ c) -4 d)none of these

Formulae-

f(x) ex-1 = loge = logee =1

x →0 x

f(x) ax-1 = loga = logea

x →0 x

f(x) log(1+x) = loge = 1

x →0 x

f(x) loga(1+x) = logae

x →0 x

f(x) log(1+1/x) = loge = 1

x →∞ 1/x

f(x) (1+1/x )x=e

x →∞

f(x) (1+x )1/x=e

x →∞

Formula based Problem-

lim (e2x-1)/x is equal to

x →0

a)1/2

(b) 2

(c) e5

(d) none of these.

Lim x→0 (2e1/x -3x)/(e1/x +x)=-

a)-3

b)0

c)2

d)4

lim (5x +3x -2)/x will be equal to

x →0

a) loge15 b) log (1/15) c) log e d) none of these

Odd & Even Function-

if f(x) is an odd function then

a){f(-x)+f(x)}/2 is an even function

b){|x|+1} is even when [x] = the integer x≤

c) [f(x)+f(-x)]/2 is neither even or odd

d) none of these

Continuity-

When a function is discontinuous-

A function will be discontinuous when the function is indeterminate.

The points of discontinuity of the function :

f(x) = (2x2+6x-5)/(12x2+x-20) are, when x is

a)-4/5 and 5/3

b)-4/3 and 5/4

c)4/5 and -5/3

d)4/3 and -5/4

Checking Limit Exist or Not-

If f(x)=(x+1)/√(6x2+3)+3x then lim f(x) and f(-1)

x →-1

a) both exists

b) one exists and other does not exist

c) both do not exists

d)none of these

lim y→0 (3y+ |y|)/(7y -5|y|)=

a)2

b)⅙

c)3/7

d)does not exist

Checking Continuity-

A function f(x) is defined as follows

f(x) = x2 when 0 < x <1

= x when 1 <_ x < 2

= (1/4) x3 when 2 < x < 3

Now f(x) is continuous at

a) x = 1 b) x = 3 c) x = 0 d) none of these.

Conclusion-

What are Limits?

Basic of Limit

Avoid Indeterminant

Infinity Problem

Formula based Problem

Condition for discontinuous

Check Limit Exist or Not

Check Continuous or Not

## Notes

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