Find the limit of (x^2 - 3x + 2) / (root(5 - x) - root(x + 1)) as x approaches 2.

Step-by-step solution

1. 1

   Confirm the indeterminate form. Substituting $x=2$: the numerator is $4-6+2=0$, and the denominator is $\sqrt{3}-\sqrt{3}=0$. So the expression is of type $\tfrac00$ — no value yet, but the shared zero means a common factor of $(x-2)$ is hiding in both parts, waiting to be cancelled.

2. 2

   Factor the numerator. Two numbers with product $2$ and sum $-3$ are $-1$ and $-2$:

   $x^{2}-3x+2=(x-1)(x-2)$

3. 3

   Rationalise the denominator with its conjugate. Multiply numerator and denominator by $\sqrt{5-x}+\sqrt{x+1}$, which turns the difference of roots into a difference of squares:

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

   The roots vanish entirely, leaving a linear expression that visibly contains the troublesome zero: $4-2x=-2(x-2)$.

4. 4

   Cancel the common factor. The expression becomes

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

   valid for $x\neq2$ — which is all a limit ever needs, since it describes behaviour near the point, not at it. This algebraic form was checked against the original at $x=1.5,1.9,2.1,2.5$ to $10^{-9}$ ✓.

5. 5

   Substitute directly now that the singularity is gone. The simplified function is continuous at $x=2$:

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

6. 6

   Verify numerically from both sides. At $x=2\pm10^{-7}$ the original quotient evaluates to $-1.73205098$ and $-1.73205063$, bracketing $-\sqrt{3}=-1.7320508$ ✓. The two-sided agreement confirms the limit exists and equals

   $-\sqrt{3}\approx-1.7321$