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$$\lim \limits_{x \to 1} \frac{\sqrt{x+3} – 2}{x-1}$$

My initial thought is to multiply by the conjugate of the numerator. Answer should be $\frac{1}{4}$, but I’m getting $$\frac{x-1}{(x-1)(\sqrt{x+3}-2)}$$.

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For the numerator, you should have gotten $$(\sqrt{x+3}-2)(\sqrt{x+3}+2) = (\sqrt{x+3})^2-2^2 = x+3-4 = x-1.$$

Note that

$$\lim \limits_{x \to 1} \frac{\sqrt{x+3} – 2}{x-1}=f'(1)$$

by the definition of the derivative, where $f(x)=\sqrt{x+3}$. Since $f'(x)=\frac{1}{2\sqrt{x+3}}$, then the limit equals $f'(1)= \frac{1}{4}$.

Set $\sqrt{x+3}-2=y\implies x=(y+2)^2-3$

$$\lim_{x\to1}\frac{\sqrt{x+3} – 2}{x-1}=\lim_{y\to0}\dfrac y{y(y+4)}=?$$

Cancel out $y$ as $y\ne0$ as $y\to0$

Since you already received good solutions, let me show that you could do a bit more and get more information.

Considering $$A= \frac{\sqrt{x+3} – 2}{x-1}$$ start setting $x=y+1$ to make $$A= \frac{\sqrt{y+4} – 2}{y}=2\frac{\sqrt{1+\frac y4} – 1}{y}$$ and consider the Taylor series $$\sqrt{1+z}=1+\frac{z}{2}-\frac{z^2}{8}+O\left(z^3\right)$$ Replace $z$ by $\frac y4$ to get $$A=2\frac{1+\frac{y}{8}-\frac{y^2}{128}+O\left(y^3\right) – 1}{y}=\frac{1}{4}-\frac{y}{64}+O\left(y^2\right)=\frac{1}{4}-\frac{x-1}{64}+O\left((x-1)^2\right)$$ which shows the limit and also how it is approached when $x\to 1$.

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