1getting same ans as celestine . use graphs sab aasan ho jayega
6th me direct value put karna
26 Answers
106Q1. lim(x→2-) (x-3)/(x2-4)
as x → 2- i.e. it is less than 2 so, x-3 <0 and x2-4<0
further x-3→ -1 and x2-4 → 0- So, if u divide u'll get -1/0- (both negative) hence lim = +∞
1hey ans are
1 ∞ 2 -∞ 3 ∞ 4 ∞ 5 -∞ 6 6
i think this as the answer wats the correct one
1
106mani. limit will exist. he has specified -2+ or -8+ or 0+ or 0-
11sorry guys
god work asish
but i am afraid u r wasting ur time as IIT or AIEEE people will never fight over ifinity
The answer to ur questions ANAND is the limit does not exist
1arey it is -a+ means the right hand limit at -a
and vice versa
106now even i have got confused
what is -a+ ?
is it -(a+) i.e. if a = 2 is it -2.00000000001
or is it (-a)+ i.e. if a=2 is it -1.9999999999
1in 3rd one
if we take lim(h-0)f(-pi/2+h)..........we get infinity.................[11]
plz tell mistake
1hey in 1st ne i m doing like this:
taking lim(h- 0) f(2-h)............and getting -infinity............[11]
106Q5. lim(x→ 0-) 1+cosecx
As x is in 4th quadrant and tending to zero, hence cosecx tends to -∞ hence the limit is 1+(-∞) = -∞
Q6. if f(x)=3(x+1),x>0
lim(x→1) f(x) = lim(x→1) 3(x+1) = 3(1+1) = 6
106Q2. lim(x→ -8+) 2x/(x+8)
here denominator tends to 0+ (i.e. positive) as numerator is negative tending to -16 so the limit is -∞
Q3. lim(x→ -π/2+) secx
sec(-π/2+) +∞ as the angle comes is in the 4th quadrant
Q4. lim(x→ 0-) 2-cotx
as x is in 4th quadrant and tending to zero, hence cotx = -∞
So, the limit is 2-(-∞) = 2+∞ = ∞
1hei in first one apply componendo and dividendo
and then put the limit
1johncena how cud u getminus infinite
ull get minus in both numerator n denominator
1yaar for sure 1) ka toh limit doesn't exist
x cann't be equal to 2....so f(2) does n't exist......
1@#7 AND #9
ANS. GIVEN DIFFERS
I M GETTING 1.-INFINITY 2.-INFINITY
