I try to calculate $\lim \limits_{x \to 0}{\frac{\sqrt{1 + x + x^2} - 1}{x}}$. I've got $\frac{\sqrt{1 + x + x^2} - 1}{x} = \sqrt{\frac{1}{x^2} + \frac{1}{x} + x} - \frac{1}{x}$ but I don't know what to do next.
Asked
Active
Viewed 71 times
2
-
you can throw away that $x^2$ in the square root in comparison to $x.$ – abel Jan 13 '15 at 14:50
4 Answers
1
Hint: $\dfrac{\sqrt{1 + x+ x^2} -1}{x} = \dfrac{x + x^2}{x (\sqrt{1 + x+ x^2} + 1)} = \dfrac{1+x}{\sqrt{1 + x+ x^2} + 1}$
Krish
- 7,102
0
$$\sqrt{1 + x + x^2} = 1 + \dfrac{1}{2}(x+x^2) + \cdots = 1 + \dfrac{1}{2}x + \cdots$$ so the $$\lim \limits_{x \to 0}{\frac{\sqrt{1 + x + x^2} - 1}{x}} = \lim \limits_{x \to 0}\dfrac{1 + 1/2 x + \cdots - 1}{x} = \dfrac{1}{2}$$
abel
- 29,170
0
Here are the steps $$ \lim \limits_{x \to 0}\left[{\frac{\sqrt{1 + x + x^2} - 1}{x}}\right] $$ $$ =\lim \limits_{x \to 0}\left[{\frac{\sqrt{1 + x + x^2} - 1}{x}}\right] \left[{\frac{\sqrt{1 + x + x^2} + 1}{\sqrt{1 + x + x^2} + 1}}\right] $$ $$ =\lim \limits_{x \to 0}\left[{\frac{1 + x + x^2 - 1}{ x\left(\sqrt{1 + x + x^2} + 1\right)}}\right] $$ $$ =\lim \limits_{x \to 0}\left[{\frac{x\left(1+ x\right)}{x\left(\sqrt{1 + x + x^2} + 1\right)}}\right] $$ $$ =\lim \limits_{x \to 0}\left[{\frac{1+ x}{\sqrt{1 + x + x^2} + 1}}\right] $$ $$ ={\frac{1+ 0}{\sqrt{1 + 0+ 0} + 1}} $$ $$ ={\frac{1}{\sqrt{1} + 1}} $$ $$ =\frac{1}{2} $$
k170
- 9,045