lim(sqrt(4x^2-x)-sqrt(9x^2+1)+x),x->infinity

$\displaystyle{\lim_{x\to \infty}}\sqrt{4x^2-x}-\sqrt{9x^2+1}+x$

$\displaystyle{\lim_{x\to \infty}}x\left(\sqrt{4-\frac 1 x}-\sqrt{3+\frac 1 x}+1\right)$

$\mbox{In the limit we can ignore the }\dfrac 1 x \mbox{ terms}$

$\displaystyle{\lim_{x\to \infty}}x\left(\sqrt{4}-\sqrt{3}+1\right)=\displaystyle{\lim_{x\to \infty}}x\left(3-\sqrt{3}\right)$

$3 > \sqrt{3} \mbox{ so } \\ \displaystyle{\lim_{x\to \infty}}x\left(3-\sqrt{3}\right)=+\infty \\ \mbox{and so } \\ \displaystyle{\lim_{x\to \infty}}\sqrt{4x^2-x}-\sqrt{9x^2+1}+x=+\infty$

lim(sqrt(4x^2-x)-sqrt(9x^2+1)+x),x->infinity 

lim_(x->infinity) (sqrt(4 x^2-x)-sqrt(9 x^2+1)+x) = -1/4

Find the following limit:
lim_(x->infinity) (x+sqrt(4 x^2-x)-sqrt(9 x^2+1))

x+sqrt(4 x^2-x)-sqrt(9 x^2+1)  =  x+sqrt(x (4 x-1))-sqrt(9 x^2+1):
lim_(x->infinity) x+sqrt(x (4 x-1))-sqrt(9 x^2+1)

The limit of x+sqrt(x (4 x-1))-sqrt(9 x^2+1) as x approaches infinity is -1/4:
-1/4

-1/4 = -1/4:
Answer: | 
| -1/4

Have a look at the graph, here........https://www.desmos.com/calculator/wslbve0v6i

It does appear that this approaches -1/4 as x→ infinity