What is the formal way find $a$ such that $\lim_{x\to4} \frac{ax-\sqrt{x}+6}{x-4} = \frac{3}{4}$?

Question

So I got this problem

Determine the value of $a$ that satisfies the following limit $$\lim_{x\to4} \frac{ax-\sqrt{x}+6}{x-4} = \frac{3}{4}$$

If we substitute $x = 4$, the value of the limit will be $\frac{4a + 4}{0}$ which is undefined. So, the numerator must have the same root as the denominator. With $x - 4 = (\sqrt{x} - 2)(\sqrt{x} + 2)$, and the one which make the limit undefined is $(\sqrt{x} - 2)$. From here, I conclude that the numerator must have ($\sqrt{x} - 2$) as a root. And from here, I can do polynomial division and got that $a = -1$.

The other way is by argue that "if the limit have a value, then make the result indeterminate form: $\frac{0}{0}$. Then, we'll get $4a + 4 = 0$ and $a = -1$.

The problem is, I find my method is not good enough for essay problems.

Answer 1

If $\lim_{x\to a}f(x)$ exists (in $\Bbb R$) but is non-zero, and $\lim_{x\to a}g(x)=0$, then $$ \lim_{x\to a}\frac{f(x)}{g(x)}\quad\text{does not exist.} $$ Therefore, if $\lim_{x\to 4}\frac{ax-\sqrt{x}+6}{x-4}$ exists, then it must be the case that $\lim_{x\to 4}ax-\sqrt{x}+6=0$. Since the function $x\mapsto ax-\sqrt{x}+6$ is elementary, it is continuous, and so we can evaluate $\lim_{x\to4}ax-\sqrt{x}+6$ by plugging in $x=4$. We find that $\lim_{x\to4}ax-\sqrt{x}+6=4a-\sqrt{4}+6=4a+4$. Hence, $a=-1$.

However, if $a=-1$, then the limit actually equals $-5/4$. To prove this, note that for $x\ge0,x\neq4$, we have $$ \frac{-x-\sqrt{x}+6}{x-4}=-\frac{x+\sqrt{x}-6}{x-4}=-\frac{(\sqrt{x}+3)(\sqrt{x}-2)}{(\sqrt{x}+2)(\sqrt{x}-2)}=-\frac{\sqrt{x}+3}{\sqrt{x}+2} \, . $$ Therefore, there is no value of $a$ for which the limit equals $3/4$.

Answer 2

The denominator tends to $0$ and the numerator tends to $4a+4$. If $4a+4 \neq 0$ then the left hand and right hand limits are not equal and the ratio has no limit. Hence we must have $a=-1$.

If $4a+4 >0$ then the right hand limit is $\infty$ and the left hand limit is $-\infty$. If $4a+4 <0$ then the right hand limit is $-\infty$ and the left hand limit is $\infty$.

When $a=-1$ we can apply L'Hopital's Rule to show that the limit is actually $-\frac 5 4$.

Answer 3

By $y=x-4 \to 0$ we have

$$\lim_{x\to4} \frac{ax-\sqrt{x}+6}{x-4}=\lim_{y\to 0} \frac{a(y+4)-\sqrt{y+4}+6}{y}=\lim_{y\to 0} \frac{ay+4a-\sqrt{y+4}+6}{y}$$

and to guarantee that limit exists we need

$$4a-\sqrt{y+4}+6 \to 0 \iff a=-1$$

and for $a=-1$

$$ \frac{-y-\sqrt{y+4}+2}{y}=-1-\frac{\sqrt{y+4}-2}{y} \to -1-\frac14=-\frac 5 4$$

indeed by $f(y)=\sqrt{y+4} \implies f'(y)=\frac1{2\sqrt{y+4}}$ we have that by the definition of derivative

$$\lim_{y\to 0}\frac{\sqrt{y+4}-2}{y}=f'(0)=\frac14$$