TITLE: Solve a limit with radicals
QUESTION [5 upvotes]: 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!!

REPLY [4 votes]: 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 :)