Find the integer value of a, b such that $\lim_{x \rightarrow 4} \frac{ax-\sqrt{x}+b}{x-4}=\frac{3}{4}$ without using the L'Hôpital's rule.

Question

Find the integer value of $a$, $b$ such that $\lim\limits_{x \rightarrow 4} \frac{ax-\sqrt{x}+b}{x-4}=\frac{3}{4}$ without using the L'Hôpital's rule.
My work:

I multiplied with $\frac{ax+\sqrt{x}+b}{ax+\sqrt{x}+b}$, and obtained:

$=\lim\limits_{x \rightarrow 4} \frac{(ax+b)^2-x}{(x-4)(ax+b+\sqrt{x})}$

$=\lim\limits_{x \rightarrow 4} \frac{(a^2x^2+2abx+b^2-x)}{(x-4)(ax+b+\sqrt{x})}$

I factored $(x-4)$ from the numerator:

$(x-4)(a^2x+2ab-1+4a^2)+(b^2+8ab-4+16a^2)$

We have the first equation

$(b^2+8ab-4+16a^2)=0$

$=\lim\limits_{x \rightarrow 4} \frac{(x-4)(a^2x+2ab-1+4a^2)}{(x-4)(ax+b+\sqrt{x})}$ $=\frac{4a^2+2ab-1+4a^2}{4a+b+2}=\frac{3}{4}$

So, the second equation is $16a^2+8ab-4+16a^2=12a+3b+6$
$32a^2-12a+8ab-3b-10=0$
But I can't find the integer value of $a$, $b$.

Answer 1

Substitute $\sqrt x=y$, so that

$$\lim_{x\to4}\frac{ax-\sqrt x+b}{x-4}=\lim_{y\to2}\frac{ay^2-y+b}{y^2-4}=\frac34$$

We have

$$\frac{ay^2-y+b}{y^2-4}=a-\frac{y-4a-b}{(y-2)(y+2)}$$

For the limit to exist, $y=2$ must be a removable singularity, so $y-2$ must divide $y+4a+b$. We have

$$\frac{y-4a-b}{y-2}=1-\frac{4a+b-2}{y-2}$$

and the remainder should be such that $4a+b-2=0$. So,

$$\lim_{y\to2}\frac{ay^2-y+b}{y^2-4}=\lim_{y\to2}\left(a-\frac1{y+2}-\frac{4a+b-2}{(y+2)(y-2)}\right)=a-\frac14=\frac34$$

Then $a=1$ and $b=-2$.

Answer 2

Note

$$\frac{ax-\sqrt{x}+b}{x-4} = a - \frac{1}{\sqrt x+2} + \frac{4a+b-2}{x-4} $$ Then, $\lim_{x \rightarrow 4} \frac{ax-\sqrt{x}+b}{x-4}=\frac{3}{4}$ requires $$a-\frac1{\sqrt4+2}= \frac34,\>\>\>\>\>4a+b-2=0$$ which yields $a=1$ and $b=-2$.

Answer 3

You were almost there! $$ \begin{cases} b^2+8ab-4+16a^2=0\\ 32a^2-12a+8ab-3b-10=0\\ \end{cases} $$ gives two solutions$^*$

$a= -\frac{1}{4},b= -1$ discarded because there are fractions

$a= 1, b = -2$, the solution you were looking for.

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$^*$

from the first equation $(b+4a)^2=4\to b+4a=\pm 2\to b=-4a\pm 2$

let $b=-4a+2$ and plug in the second equation

$32a^2-12a+8a(-4a+2)-3(-4a+2)-10=0\to 16 (a-1)=0\to a=1$

and $b = -2$

the other solution $b=-4a-2$ leads to the fractions.

Answer 4

$(2-\sqrt{x})(2+\sqrt{x})=4-x$ is a great start for partial fraction decomposition.

$a+\frac{1}{ 2 + \sqrt{x}}+\frac{-2 + 4 a + b}{-2 + \sqrt{x}}=\frac{a(x-4)+B(2-\sqrt{x})+C}{x-4}$

with $B=-1$ and $C=-2+4a+b$.

Take into account that

$$\frac{a x- \sqrt{x} + b}{x - 4}=a-\frac{1}{ 2 + \sqrt{x}}+\frac{-2 + 4 a +b}{-2 + \sqrt{x}}$$

This denominator is not compensable anymore by $\sqrt{x}$ in the nominator. So the nominators have to be zero for the existence of the limit.

The last summand has to be eliminated for a limit to exist!

$$\frac{4 a + b-2}{\sqrt{x}-2}$$

That can only the satisfied once for $-2$ or for $y$ but not for both to not singular.

We have $\frac{3}{4}=a-\frac{1}{4}$ and $4a+b=2$ so that $a=1$ so $b=-2$.

The back substition show we only have reals and therefore only are in need of the positive solution at $x=4$. The calculation remains valid.

The solution is for fitting to the given value/limit:

$$\frac{ x - \sqrt{x} -2}{x - 4}$$.

At $x=4$ the nominator has a zero $ 4 - \sqrt{4} -2=0$ that compensates for the zero of the polynomial of first order in $x$ in the denominator. The nominator is not a polynomial in $x$ but a sum of a polynomial and a second order root. Both a steady and differentiable in the singularity of the denominator.

A plot shows the singularity is gone:

The solution is $\frac{1}{2}$ for $x=0$ and $1$ for $x\rightarrow\infty$ and $\frac{3}{4}$ at $x=4 and this is no limit!$.

The solution process is substition of the root. Then partial fraction decompostion and the fitting the parameters of the given functions to the singularity at $x=4$. Finally we analysed the function and displayed it after resubstitution.

We needed a factor in addition to match the given limit. The pair $(a,b)=(1,-2)$ is unique.