limits

u cannot explain these limits questions as better as u can do.
everyone has his/her own methods of doin it.

then i dont hav any idea

sahi pakda boss

ab tu bata de
isse kaise karein[17]

koi to isse kar do

yy is this thread neglected[2]

1-³√2

sir kar ke bata do na

meko taylor series nahin aat[2]

THE ANSWER IS 2/9
I HOPE THIS WOULD HELP

wat r u doin manas??

i think this is lhospital....

L=lim 2x-³âˆšx³+x²+1-³âˆšx³-x²+1
1/x²

now use lhosiptal......

divide & multiply wid x den it wud b ∞/∞ form den apply L'hospital guess dats gonna help

akand ne yehi kia hai
uss se phas rahe hain[2]

First, set y=1/x, so that the required limit becomes
\lim_{y\to 0} \dfrac{2-\sqrt[3]{y^3+y+1}-\sqrt[3]{y^3-y+1}}{y^2}
Since y→0, therefore y³+y+1≈1+y and y³-y+1≈1-y. So the above limit becomes
\lim_{y\to 0} \dfrac{2-\sqrt[3]{1+y}-\sqrt[3]{1-y}}{y^2}
Apply the binomial expansions
\sqrt[3]{1+y}=1+\dfrac{y}{3}-\dfrac{1}{9}y^2+\dfrac{10}{3^3}y^3+\ldots + terms containing y⁴ and higher powers.
and
\sqrt[3]{1-y}=1-\dfrac{y}{3}-\dfrac{1}{9}y^2-\dfrac{10}{3^3}y^3-\ldots-terms containing y⁴ and higher powers.
Therefore,
2-³√1+y-³√1-y=2y²/9 + terms containing y⁴ and higher powers.
So when we divide by y², we get the required ratio as
2/9 + terms containing y² and higher powers of y.
Hence when y→0, we get the required limit as 2/9.

yah but ultimately when u get 0 other things dont matter

na ji na
keep trying

hey divide the whole equation by x² and then take the remaining x inside the cube root then it will become x³.

then u can divide the eqn under the root and then u can put x=infinity and then u get

2-1-1=0 is the ans

ALARM

WARNING : MANAS U R IN GR8 TROUBLE

DUDE HOW CAN U DIVIDE AN EQUATION BY x WHEN x TENDS TO INFINTY ????

then wats the ans for the question

hey u can .
u do the things in that way only

bhai the mistake u r committing is the following

IN MATHEMATICS
U CAN NEVER DIVIDE AN EQUATION WITH A NUMBER 0 OR ∞ AS IT MAKES NO SENSE[1]

ans 0