limit ${\lim_{x \to 49} \frac{\sqrt{x}-7}{x-49} }.$ I have to find the limit
$${\lim_{x \to 49} \frac{\sqrt{x}-7}{x-49} }.$$
I know that I cannot plug in $49$ because that would make the denominator $0$. I was told to rationalize the numerator and I did. This is what I did but I got the incorrect answer:
$$\dfrac{\sqrt{x}-7}{x-49}\times\dfrac{\sqrt{x}+7}{\sqrt{x}+7}.$$
I multiplied this out and got $$\dfrac{x-49}{x\sqrt{x}+7x-49\sqrt{x}+343}.$$ Now when I plugged in $49$, the limit came out as $0$ but it was incorrect. Am I missing a step or did I do something wrong? I went over it a couple of times and I cannot catch my mistake.
 A: Factorise $x-49$ as $(\sqrt x+7)(\sqrt x-7)$, and divide the numerator and denominator by $\sqrt{x}-7$.
A: Let $\sqrt{x}=y$. We are then looking at $\displaystyle\lim_{y\to 7}\frac{y-7}{y^2-49}$. Familiar?
A: $$\begin{align}
\lim_{x\to 49}\frac{\sqrt{x}-7}{x-49}=&\lim_{x \to 49}\frac{(\sqrt{x}-7)}{x-49}\frac{(\sqrt{x}+7)}{(\sqrt{x}+7)}\\
&=\lim_{x \to 49}\frac{(\sqrt{x})^2-7^2}{x-49}\frac{1}{(\sqrt{x}+7)}\\
&=\lim_{x \to 49}\frac{x-49}{x-49}\frac{1}{(\sqrt{x}+7)}\\
&=\lim_{x \to 49}\frac{1}{\sqrt{x}+7}\\
&=\frac{1}{14}
\end{align}$$
A: If you've learned L'Hôpital's rule, you know that this reduces to:
$$\lim_{x\to49}\frac{1}{2\sqrt{x}} = \frac{1}{14}$$
A: We know that the difference between two squares is  $x^2-a^2=(x-a)(x+a)$ so by writting $x-49$ as a difference between two squares we get$$x-49=(\sqrt{x}^2-7^2)=(\sqrt{x}-7)(\sqrt{x}+7)$$ which implies that $$\lim_{x\rightarrow 49} \frac{\sqrt{x}-7}{(\sqrt{x}-7)(\sqrt{x}+7)}=\lim_{x\rightarrow 49} \frac{1}{(\sqrt{x}+7)}=\frac{1}{14}.$$
A: So close! You actually did too much work. If you leave the denominator as-is you get
$$
\dfrac{x-49}{(x-49)(\sqrt{x}+7)} = \dfrac{1}{\sqrt{x}+7}
$$
which you can then take the limit of.
