limits

Q1. 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 = +∞

hey ans are

1 ∞ 2 -∞ 3 ∞ 4 ∞ 5 -∞ 6 6
i think this as the answer wats the correct one

getting same ans as celestine . use graphs sab aasan ho jayega

6th me direct value put karna

ans are

1 ∞
2 -∞
3 ∞
4 ∞
5 -∞
6 6

mani. limit will exist. he has specified -2+ or -8+ or 0+ or 0-

sorry 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

arey it is -a+ means the right hand limit at -a
and vice versa

now 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

in 3rd one

if we take lim(h-0)f(-pi/2+h)..........we get infinity.................[11]

plz tell mistake

3rd one ans. given is -infinity

Q5. 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

Q2. 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+∞ = ∞

hei in first one apply componendo and dividendo
and then put the limit

VECTOR POST UR SOLUTION TO CLARIFY WAT DO U MEAN!

johncena how cud u getminus infinite
ull get minus in both numerator n denominator

yaar for sure 1) ka toh limit doesn't exist

x cann't be equal to 2....so f(2) does n't exist......

@#7 AND #9

ANS. GIVEN DIFFERS

I M GETTING 1.-INFINITY 2.-INFINITY

2^nd is -16

if f(x)=3(x+1),x>0

find \lim_{x \to 1}f(x)

\lim_{x \to 0-}2-cotx

\lim_{x \to -pi/2+}secx