Find $\lim_{x \to 4}\frac{x^2-5x+4}{\sqrt{x}-2}$ without L'Hôpital's rule

Question

I am learning limits now and I have a problem:

Find $\lim_{x \to 4} \frac{x^2-5x+4}{\sqrt{x}-2}$

I have tried to rationalize the denominator and I have got something like this:

$\lim_{x \to 4} \frac{(x-4)(x-1)(\sqrt{x}+2)}{x-4}$

I don't know how to continue from here, can you give me a tip or show me a solution to this?

Answer 1

Tip:

Check your factorization of $x^2-5x+4$.

---

Another tip now that you corrected the factorization:

cancel $x-4$ from numerator and denominator.

Answer 2

One thing that comes to my mind is to let $x = y^2$ (so if $y \to 2$ then $x \to 4$). Then

$$\frac{x^2-5x+4}{\sqrt{x}-2} = \frac{y^4 - 5y^2 + 4}{y - 2}.$$

Now you can use long division or synthetic division to get

$$ \frac{y^4 - 5y^2 + 4}{y - 2} = y^3 + 2y^2 - y - 2, \quad (y\neq 2). $$

Answer 3

$$\lim_{x \to 4} \frac{x^2-5x+4}{\sqrt{x}-2}=\lim_{x \to 4}\frac{(x-4)(x-1)}{\sqrt{x}-2}=\lim_{x \to 4}\frac{(\sqrt x-2)(\sqrt x+2) (x-1)}{\sqrt{x}-2}$$

$$=\lim_{x \to 4}(\sqrt x+2) (x-1)=4\times 3=12$$