Xjyuk

The value of the $$limlimits_xto-infty(4x^2-x)^1/2 +2x$$ is?

The answer given is $1/4$.

I rationalized and got $$limlimits_xto-infty frac -xx$$ how to proceed further?

The limit is equivalent to
$$beginalign
lim_xto-inftyleft(2|x|left(1-frac14xright)^1/2+2xright)
&=lim_xto-inftyleft(-2xleft(1-frac14xright)^1/2+2xright)\
&=lim_xto-infty-2xleft(left(1-frac14xright)^1/2-1right)\
endalign$$
Then using the generalized binomial expansion we get that as $xto0$
$$(1+x)^n=1+nx+o(x)$$
Hence our limit becomes
$$beginalign
lim_xto-infty-2xleft(left(1-frac14xright)^1/2-1right)
&=lim_xto-infty-2xleft(1-frac18x+oleft(frac1xright)-1right)\
&=lim_xto-infty-2xleft(-frac18x+oleft(frac1xright)right)\
&=lim_xto-inftyleft(frac14+o(1)right)\
&=frac14\
endalign$$

A more elemental solution :
$$lim_xto -infty big( sqrt4x^2-x +2x big) =lim_xto -infty frac-xsqrt4x^2-x -2x =lim_xto infty fracxsqrt4x^2+x+2x =lim_xtoinfty frac1sqrt4+frac1x+2 = frac1sqrt4+2=frac14$$
where I used that
$$lim_xto infty f(x)=lim_xto -infty f(-x)$$

beginalign*
(4x^2-x)^1/2+2x
&=sqrt4x^2-x+2x
\&=frac(sqrt4x^2-x+2x)(sqrt4x^2-x-2x)sqrt4x^2-x-2x
\&=frac(4x^2-x)-4x^2sqrt4x^2-x-2x
\&=frac4x^2-x-4x^2sqrt4x^2-x-2x
\&=frac-xsqrt4x^2-x-2x
\&=frac-xsqrt(2x)^2(1-frac14x)-2x
\&=frac-xx
\&qquad [x<0]
\&=frac-x-2xsqrt1-frac14x-2x
\&=frac12sqrt1-frac14x+2
\&tofrac12sqrt1+0+2
\&=frac14.
endalign*
(as $xto-infty$)

Your rationalization is almost correct. It should be

$$frac-xsqrt4x^2-x-2x = frac-xleft[left(4-frac1xright)^1/2-frac2xright].$$

You have assumed that $-2x = -2|x|,$ which is not true when $x$ is negative. But as $xto-infty,$ we can asume $x<0$ and thus $|x|=-x,$ so this is equal to

$$frac1left(4-frac1xright)^1/2+2.$$

For the sake of comfort, we change the sign and evaluate

$$limlimits_xtoinfty(4x^2+x)^1/2-2x$$

which is

$$limlimits_xtoinftyfracx(4x^2+x)^1/2+2x$$

or

$$limlimits_xtoinftyfrac1(4+frac1x)^1/2+2=frac14.$$

$y:=-x$, and $ lim y rightarrow +infty$.

$(4y^2+y)^1/2-2y=$

$((2y+1/4)^2-1/16)^1/2-2y$;

$z:=2y+1/4$;

We get

$((z^2-1/16)^1/2-z) +1/4$;

Since

$lim_z rightarrow infty (z^2-1/16)^1/2-z)=0$ (why?), and we are done.

Note:

$(z^2-1/16)^1/2-(z^2)^1/2= $

$dfrac-1/16(z^2-1/16)^1/2+(z^2)^1/2$

Hint: Multiply by $sqrt4x^2-x-2xover sqrt4x^2-x-2x$

The result is $-xoversqrt4x^2-x-2x$

$=-xover x=$

$-xover -xsqrt4-1over x+2$

$1over sqrt4-1over x+2$.