Question

plz give the answer​

Answer 1

Answer:

1 / 2

Step-by-step explanation:

Use L'Hopital's Rule since the expression tends to the form 0 / 0:

$\displaystyle \lim_{x\rightarrow0}\frac{(1+x+x^2)^{1/2}-1}{x}\\= \lim_{x\rightarrow0}\frac{\tfrac12(1+x+x^2)^{-1/2}(1+2x) - 0}{1}\\=\lim_{x\rightarrow0}\frac{1+2x}{2\sqrt{1+x+x^2}}\\=\frac{1}{2}$

Alternatively, use the power series expansion for the square root (it's just a generalized binomial expansion):

$\displaystyle\sqrt{1+x} = 1 + \left(\tfrac12\right)x + \left(\tfrac{1}{2}\right)\left(\tfrac{-1}{2}\right)x^2 + \left(\tfrac{1}{2}\right)\left(\tfrac{-1}{2}\right)\left(\tfrac{-3}{2}\right)x^3+\cdots$

Then

$\displaystyle\frac{\sqrt{1+x+x^2}-1}{x}\\=\tfrac12(1+x) -\tfrac14(1+x)(x+x^2) + \tfrac38(1+x)(x+x^2)^2 -\cdots\\\longrightarrow \tfrac12, \ \text{as}\ x\rightarrow 0.$