Vtyjkyuk

Find the limit: $$ lim_x rightarrow + infty(sqrtx^2 + 2x - sqrtx^2 - 7x)$$

I did the following:

beginalign
(sqrtx^2 + 2x - sqrtx^2 - 7x) = frac(sqrtx^2 + 2x - sqrtx^2 - 7x)1 cdot frac(sqrtx^2 + 2x + sqrtx^2 - 7x)(sqrtx^2 + 2x + sqrtx^2 - 7x)
endalign

I know the final answer is $frac92$. After multiplying by the conjugate, I see where the $9$ in the numerator comes from. I just can't remember how I solved the rest of the problem.

$$lim_xto inftyfrac(sqrtx^2+2x-sqrtx^2-7x)(sqrtx^2+2x+sqrtx^2-7x)sqrtx^2+2x+sqrtx^2-7x$$
$$=lim_xto inftyfrac9xsqrtx^2+2x+sqrtx^2-7x=lim_xto inftyfracfrac9xxfracsqrtx^2+2xx+fracsqrtx^2-7xx$$$$=lim_xto inftyfrac9sqrtfracx^2+2xx^2+sqrtfracx^2-7xx^2=lim_xto inftyfrac9sqrt1+frac 2x+sqrt1-frac 7x=frac91+1=frac 92.$$

Expanding the rightmost term, we get:
$$
frac(x^2+2x) - (x^2-7x)sqrtx^2+2x + sqrtx^2-7x = frac9xsqrtx^2+2x + sqrtx^2-7x
$$
now just divide the numerator and denominator by $x = sqrtx^2$:
$$
frac9sqrt1+frac2x + sqrt1-frac7x.
$$
We have that $sqrt1+frac2x + sqrt1-frac7x rightarrow 2$ as $x rightarrow infty$, so the final answer is $9/2$.

Setting $dfrac1x=h,$ the limit reduces to $$lim_hto0^+fracsqrt1+2h-sqrt1-7hh$$

$$=lim_hto0^+frac1+2h-(1-7h)h(sqrt1+2h+sqrt1-7h)$$

$$=lim_hto0^+frac9sqrt1+2h+sqrt1-7h$$

$$=frac9sqrt1+2cdot0+sqrt1-7cdot0$$

$bfMy; Solution::$ Given $displaystyle lim_xrightarrow inftyleft(sqrtx^2+2x-sqrtx^2-7xright) = lim_xrightarrow inftyxleftleft(1+frac2xright)^frac12-left(1-frac7xright)^frac12right$

Now Using Binomial Expansion::

$displaystyle lim_xrightarrow inftyleftleft(1+frac12cdot frac2x+frac12cdot -frac12cdot frac42x^2......+inftyright)-left(1-frac12cdot frac7x-frac12cdot -frac12cdot frac492x^2.+inftyright)right$

So $displaystyle lim_xrightarrow inftyleft(sqrtx^2+2x-sqrtx^2-7xright) = 1+frac72 = frac92$

$$lim_xrightarrow infty (sqrtx^2+2x - sqrtx^2-7x) $$$$=lim_xrightarrow infty bigg(sqrtx^2+2x+1 - sqrtx^2-7x+frac494bigg) $$ $$=lim_xrightarrow infty bigg( x+1 - bigg(x-frac72bigg)bigg) =frac92 $$

The mean value theorem gives $f(x+h) = f(x) + f'(x)h + 1 over 2 f''(xi) h^2$, where $xi in [x,x+h]$ (adjusted appropriately for sign of $h$).

This gives $|sqrt1+theta - (1+ 1 over 2 theta) | le 1 over 8 theta^2$.

We have (if $x>0$),$sqrtx^2+2x - sqrtx^2-7x = x(sqrt1+2 over x-sqrt1-7 over 7)$.

Then $| sqrt1+2 over x-sqrt1-7 over x - 1 over 29 over x| le 1 over 8 ((2 over x)^2+(7 over x)^2)$.

Multiplying through by $x$ and letting $x to infty$ yields the desired result.

Note:
The point of this answer is that you can guess the answer by
using $sqrt1+theta approx 1+ 1 over 2 theta$ and
writing
$sqrt1+2 over x-sqrt1-7 over x approx 1 over 29 over x$.