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CPT Chapter Limits and Continuity

CA CPT Exam  

With Anand.Duration:4hrs25min

Limits and ContinuityDuration (min:sec)
Limits 
Overview07:49
What is Limit18:21
Limit with graph03:35
Basics of Limit24:27
Avoid Indeterminant124:41
Avoid Indeterminant222:02
How to solve infinity problem02:14
Infinity Problem21:03
Limit Formulae06:19
Formulae Example116:42
Formulae Example229:09
Odd & Even Function07:07
Continuity 
Continuity Concept14:25
Function Discontinuous03:54
Checking Limit Exist Concept03:57
Example Checking Limit Exist12:58
Check Continuity-Concept02:40
Check Countinity-Example117:28
Check Countinity-Example223:18
Conclusion02:37
Total04:24:46

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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