1u cannot explain these limits questions as better as u can do.
everyone has his/her own methods of doin it.
38 Answers
1
1wat r u doin manas??
i think this is lhospital....
L=lim 2x-3√x3+x2+1-3√x3-x2+1
1/x2
now use lhosiptal......
1divide & multiply wid x den it wud b ∞/∞ form den apply L'hospital guess dats gonna help
62Using binomial expansion for fractional powers....
(1+1/x+1/x^3)^{1/3} =1+\frac{1}{3}(1/x+1/x^3)+\frac{\frac{1}{3}(\frac{1}{3}-1)}{1.2}(1/x+1/x^3)^2 + \text{higher order terms}
(1-1/x+1/x^3)^{1/3} =1+\frac{1}{3}(-1/x+1/x^3)+\frac{\frac{1}{3}(\frac{1}{3}-1)}{1.2}(-1/x+1/x^3)^2 + \text{higher order terms}
Now you can add the two and get
(1-1/x+1/x^3)^{1/3} + (1-1/x+1/x^3)^{1/3} =2+\frac{2}{3}(1/x^3)-2/9x^2+\text{higher order}
66First, 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.
62The original question now becomes
\lim_{x\rightarrow \infty} x^2\times(-\frac{2}{3}(1/x^3)+2/9x^2+\text{higher order})
Thus it becomes
\lim_{x\rightarrow \infty} (2/9-\frac{2}{3}(1/x)+\text{higher order})
= 2/9
1hey 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
11ALARM
WARNING : MANAS U R IN GR8 TROUBLE
DUDE HOW CAN U DIVIDE AN EQUATION BY x WHEN x TENDS TO INFINTY ????
11bhai 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]
62manas you can divide but dont you think you have to mulitply it back?
