Find the limit of (root(4 - x) - 3)/(x^2 + 7x + 10) as x approaches -5.

Step-by-step solution

1. 1

   Confirm the indeterminate form. At $x=-5$ the numerator is $\sqrt{5+4}-3=\sqrt9-3=0$, and the denominator is $25-35+10=0$. So the quotient is of type $\tfrac00$, meaning a factor of $(x+5)$ divides both parts and must be cancelled before substituting.

2. 2

   Factor the denominator. Two numbers with product $10$ and sum $7$ are $5$ and $2$:

   $x^{2}+7x+10=(x+5)(x+2)$

   The factor $(x+5)$ is now explicit; the job is to produce a matching one upstairs.

3. 3

   Multiply by the conjugate of the numerator. Using $\left(\sqrt{u}-3\right)\left(\sqrt{u}+3\right)=u-9$ with $u=4-x$:

   $\left(\sqrt{4-x}-3\right)\left(\sqrt{4-x}+3\right)=(4-x)-9=-x-5=-(x+5)$

   The radical disappears and the hoped-for $(x+5)$ appears — with a minus sign that must be carried.

4. 4

   Cancel and simplify. The quotient becomes

   $\frac{-(x+5)}{(x+5)(x+2)\left(\sqrt{4-x}+3\right)}=\frac{-1}{(x+2)\left(\sqrt{4-x}+3\right)}$

   valid for $x\neq-5$, which is exactly the punctured neighbourhood a limit examines. This form was checked against the original at $x=-6,-5.1,-4.9,-4$ to $10^{-8}$ ✓.

5. 5

   Substitute into the simplified expression. It is continuous at $x=-5$:

   $\frac{-1}{(-5+2)\left(\sqrt{9}+3\right)}=\frac{-1}{(-3)(6)}=\frac{1}{18}$

   The two negatives cancel, so the limit is positive.

6. 6

   Verify numerically. At $x=-5\pm10^{-5}$ the original quotient evaluates to $0.05555576$ and $0.05555535$, straddling $\tfrac1{18}=0.0555556$ ✓. Both one-sided limits agree, so the limit exists and equals $\tfrac{1}{18}$.