sign up I login
 advanced
» win an I-Phone. check i-points

Ask & Discuss Questions with Community & Experts

Moderation Team
 90 chars left    advanced
Ask iit jee aieee pet cbse icse state board community Community Discussion Question: limits of composite functions
Forum Index -> Differential Calculus like the article? email it to a friend.  
Author Message
ADPDAVCTC (14)

Cool goIITian

Olaaa!! Perrrfect answer. 2  [4 rates]

ADPDAVCTC's Avatar

total posts: 72    
offline Offline

 


 f{g(x)}=f{g(x) }=f(m),provided f is continuous at g(x)=m.


for eg:let us assume that g(x)=sgn(x)=  & f(x)= constant....


then, f{g(x)} = constanteven if g is discontinuous at x=0.....


plizzzz..... someone explain me "provided f is continuous at g(x)=m"


what is the problem if f is discontinuous?????????


i also can't understand the example.......i think its an exception to the above concept........


RATES ASSURED.........OKAYYYYYYYYY


adipic acid with silver oxide
    
ADPDAVCTC (14)

Cool goIITian

Olaaa!! Perrrfect answer. 2  [4 rates]

ADPDAVCTC's Avatar

total posts: 72    
offline Offline
plizzzzz somebody help me ...........is no one able 2 understand this discontinuity concept in limits??????????

adipic acid with silver oxide
 this reply: 0 points  (with Olaaa!! Perrrfect answer.   in 0 votes )   [?]
 
You have to be logged on to rate
  
v.sabarish (177)

Scorching goIITian

Olaaa!! Perrrfect answer. 33  [39 rates]

v.sabarish's Avatar

total posts: 277    
offline Offline
Try to think of a graphical solution...........

take, for example, f(x)=x.x
at x=1, you have the function value as 1, at x=2, it is 4


Now, if you consider f(2x), at x=1, the function value is 4, at x=2, it is 16......

Here, you perform two operations... the domain of the function f is not the region where you consider the values of x, but instead, it's the range of the function contained within f, i.e, the range of (2x) for a given interval for x in my example.....

in your question, the first part..... you are taking the limiting value of the function f..... but it's domain is not a neighbourhood of a, but instead, it's domain is the range of g, when g has the neighbourhood of a as its domain.... I guess you understand me....
as x tends to a, the value of g(x) tends to some other value, which may or maynot be equal to a, in your case, m.
 this reply: 0 points  (with Olaaa!! Perrrfect answer.   in 0 votes )   [?]
 
You have to be logged on to rate
  
v.sabarish (177)

Scorching goIITian

Olaaa!! Perrrfect answer. 33  [39 rates]

v.sabarish's Avatar

total posts: 277    
offline Offline
I've given a clear hint, you try to think of the exact relation between your question and my answer, if you don't find any, I shall be ready with a reply...
 this reply: 0 points  (with Olaaa!! Perrrfect answer.   in 0 votes )   [?]
 
You have to be logged on to rate
  
v.sabarish (177)

Scorching goIITian

Olaaa!! Perrrfect answer. 33  [39 rates]

v.sabarish's Avatar

total posts: 277    
offline Offline
Starting it all over again......, think about the graphs of the functions....

for a function f(x), if someone asks you to check the continuity, you are only worried about the 'y' values, the output values of the function and check if they are continuous in some given interval, an interval in 'x'... if this is true, you say the function is continuous, and whenever a function is continuous, you can take the limiting value as x tends to some 'a' as f(a)
Whenever the function is not continuous, two cases arise, the limit doesnot exist... i.e, the function has two curves that don't meet, like the step function or the signum function.... or even y=1/x...
else, you can have some function like y=x for all x not equal to 0 and y=1 for x=0.. here, you have a value for the limit, as x tends to 0, yet the function is discontinuous...
 this reply: 0 points  (with Olaaa!! Perrrfect answer.   in 0 votes )   [?]
 
You have to be logged on to rate
  
v.sabarish (177)

Scorching goIITian

Olaaa!! Perrrfect answer. 33  [39 rates]

v.sabarish's Avatar

total posts: 277    
offline Offline
Trace the graphs, you'll understand each part of my above explanation

Coming to what you've asked... about the continuity of 'g'...
the range of 'g' becomes the domain of 'f'... this must be very clear now....
you are concerned about the neighbourhood of x=a.. here, g(x)=m...
If a function is continuous, you can simply take the value of lim as x tends to a as f(a)...
here, take that y=g(x).. replace this y in f(g(x))
it becomes limit as x tends to a, f(g(x)), or lim as x tends to a, f(y)
but as x tends to 'a', g(x) tends to 'm', or simply, y tends to 'm'...
so, we can rewrite that as limit as y tends to m, f(y)
you first go to prove that f is continuous at y=m.. i.e, f is continuous at g(x)=m..... when this is proved, you can simply take limit as y tends to m, f(y) as f(m)......
 this reply: 5 points  (with Olaaa!! Perrrfect answer.   in 1 votes )   [?]
 
You have to be logged on to rate
  
v.sabarish (177)

Scorching goIITian

Olaaa!! Perrrfect answer. 33  [39 rates]

v.sabarish's Avatar

total posts: 277    
offline Offline
About the example... you need to understand the difference between an 'if -then' statement and an 'if and only if' statement....
let me show you something....
you divide 64 by 16....
64/16.. you cancel off the '6' in the numerator and denominator... you get 4/1, hence the answer is 4
same with 95/19, cancel both nines, answer is 5.....
This doesn't mean you can do this everytime... just because this technique works in some cases, you can't generalize this... when you generalize, you need to tell something that holds in all cases... that doesn't mean that there is no case against it...
there are many statements in math which are very true, yet their converse may not always be true...
You need to know more about statements.... try some books
 this reply: 0 points  (with Olaaa!! Perrrfect answer.   in 0 votes )   [?]
 
You have to be logged on to rate
  
 
Forum Index -> Differential Calculus
Go to:   

 Aakash Institute IIT/ AIEEE/ Medical Crash Course
Name  
E-mail  
Phone  
Mobile  
** Hurry. Exclusive goIIT Offer. Limited Seats Only!
available in: New Delhi, Amritsar, Bhatinda, Bokaro, Chandigarj, Dehradun, Guwhati, Hyderabad, Indore, Jaipur, Kanpur, Karnal, Kolkata, Kota, Lucknow, Ludhiana, Mumbai, Noida, Patiala, Patna, Pune, Ranchi, Varanasi
Top Offers for goIITians
Correspondence Courses
Brilliant Tutorials
Narayana Institute
Aakash Institute
Classroom/Crash Courses
Aakash-IITJEE : AIEEE
Aakash-IITJEE : DCE
Aakash-IITJEE : MHTCET
Aakash Institute : AIPMT
Online Test Series
Brilliant Tutorials
Narayana Institute
Aakash Institute
Mahesh Tutorials
AMITY      Sri Chaitanya