Evaluate the limit of (sqrt(3 + x) - sqrt(3))/x as x approaches 0.

Step-by-step solution

1. 1

   Confirm the indeterminate form. At $x=0$ the numerator is $\sqrt{3}-\sqrt{3}=0$ and the denominator is $0$, giving $\frac{0}{0}$. Substitution alone cannot finish the job.

2. 2

   Choose rationalisation over factoring. The obstruction is a difference of square roots, so multiply numerator and denominator by the conjugate $\sqrt{3+x}+\sqrt{3}$. This uses $(a-b)(a+b)=a^2-b^2$ to remove the radicals from the top.

3. 3

   Simplify the numerator. $\left(\sqrt{3+x}\right)^2-\left(\sqrt{3}\right)^2=(3+x)-3=x$, so the quotient becomes $\frac{x}{x\left(\sqrt{3+x}+\sqrt{3}\right)}$.

4. 4

   Cancel the factor of $x$. For $x\neq 0$ this equals $\frac{1}{\sqrt{3+x}+\sqrt{3}}$, and a limit only inspects values near $0$, never $0$ itself.

5. 5

   Substitute $x=0$. The reduced expression is continuous there, giving $\frac{1}{2\sqrt{3}}=\frac{\sqrt{3}}{6}$ after rationalising the denominator.

6. 6

   Check numerically. A difference quotient at $x=10^{-8}$ returns $0.288675$, and $\frac{\sqrt{3}}{6}=0.2886751$ — agreement to six decimals.