Solve a limit with radicals I don't know how to solve this limit. What should I do?
$$
\lim_{x\to 0} {\sqrt{x^3+2x+1}-\sqrt{x^2-3x+1} \over \sqrt{4x^2-3x+1} - \sqrt{2x^3+6x^2+5x+1}}
$$
Thank you!!
 A: Multiply the given expression by a special kind of $1$ :
$1 =  \frac{\sqrt{x^3+2x+1}+\sqrt{x^2-3x+1}}{ \sqrt{4x^2-3x+1} + \sqrt{2x^3+6x^2+5x+1}}  . \frac{ \sqrt{4x^2-3x+1} + \sqrt{2x^3+6x^2+5x+1}}{ \sqrt{x^3+2x+1}+\sqrt{x^2-3x+1}  }$

$\lim \limits_{x\to 0} \frac{\sqrt{x^3+2x+1}-\sqrt{x^2-3x+1}}{ \sqrt{4x^2-3x+1} - \sqrt{2x^3+6x^2+5x+1}} 
\\= \lim \limits_{x\to 0} {\sqrt{x^3+2x+1}-\sqrt{x^2-3x+1} \over \sqrt{4x^2-3x+1} - \sqrt{2x^3+6x^2+5x+1}} .\frac{\sqrt{x^3+2x+1}+\sqrt{x^2-3x+1}}{ \sqrt{4x^2-3x+1} + \sqrt{2x^3+6x^2+5x+1}}  .\frac{ \sqrt{4x^2-3x+1} + \sqrt{2x^3+6x^2+5x+1}}{ \sqrt{x^3+2x+1}+\sqrt{x^2-3x+1}  }
\\= \lim \limits_{x\to 0} \frac{x^3-x^2+5x}{-2x^3-2x^2-8x}  .\frac{ \sqrt{4x^2-3x+1} + \sqrt{2x^3+6x^2+5x+1}}{ \sqrt{x^3+2x+1}+\sqrt{x^2-3x+1}  }
\\= \lim \limits_{x\to 0} \frac{x^2-x+5}{-2x^2-2x-8}  .\frac{ \sqrt{4x^2-3x+1} + \sqrt{2x^3+6x^2+5x+1}}{ \sqrt{x^3+2x+1}+\sqrt{x^2-3x+1}  }
$
plugin x = 0
A: As David H commented, rationalizing the denominator would be a good starting point.
Now, if you know Taylor series, the problem starts to be simple since, around $x=0$, you have $$\sqrt{x^3+2x+1}=1+x+O\left(x^2\right)$$ $$\sqrt{x^2-3x+1}=1-\frac{3 x}{2}+O\left(x^2\right)$$ $$\sqrt{4x^2-3x+1}=1-\frac{3 x}{2}+O\left(x^2\right)$$ $$\sqrt{2x^3+6x^2+5x+1}=1+\frac{5 x}{2}+O\left(x^2\right)$$ So, numerator is $$ \frac{5 x}{2}+O\left(x^2\right)$$and denominator is $$-4 x+O\left(x^2\right)$$ 
I am sure that you can take from here.
All of the above has been done using the fact that, if $x$ is small $$\sqrt{a+bx+cx^2+dx^3+ex^4+\cdots}\simeq \sqrt{a+bx}=\sqrt{a}+\frac{b x}{2 \sqrt{a}}+O\left(x^2\right)$$
A: Well, for small numbers, 
$$
{x^3 << x^2 << x << 1}
$$
So the given integral basically reduces to,
$$
\lim_{x\to 0} {\sqrt{2x+1}-\sqrt{1-3x} \over \sqrt{1-3x} - \sqrt{5x+1}}
$$
as higher powers become negligible. This is the same as ignoring lower exponents of $x$ for limits that tend to infinity.
Recalling that for
$$
{ x << 1}
$$
we have,
$$
{(1 + x)^{1/2} = 1 + x/2 - x^2 / 8 + ...}
$$
We can similarly ignore higher exponents of x, thus we have,
$$
{(1 + x)^{1/2} \approx 1 + x/2}
$$
Thus the limit simplifies to,
$$
\lim_{x\to 0} {(1+x)  - (1- 3x/ 2 )\over (1- 3x/2) - (1 + 5x/2)}
$$
Which reduces to $-5 \over 8$$=-0.625$.
Trial value, putting $x = 0.1$, we have $-0.65017361$
for $x = 0.01$, we have $-0.62692914$, as we can see the limit converges to the required value.
Hope this helped :)
A: The expression is of the form $\frac{\sqrt a-\sqrt b}{\sqrt c-\sqrt d}$
Rationalise/simplify as $\sqrt a-\sqrt b\times \frac{\sqrt a+\sqrt b}{\sqrt a+\sqrt b}\times \frac{1}{\sqrt c-\sqrt d}\times \frac{\sqrt c+\sqrt d}{\sqrt c+\sqrt d}$
Answer is $\frac{-5}{8}$
