Evaluate the limit of (sqrt(x + 11) - 6)/(x - 5) as x approaches 5, or explain why it does not exist.

Step-by-step solution

1. 1

   Substitute before choosing a technique. The shape $\dfrac{\sqrt{u(x)}-c}{x-a}$ usually signals "rationalise the numerator", but that is only correct when the numerator actually vanishes. Check it:

   $\sqrt{5+11}-6=\sqrt{16}-6=4-6=-2$

   and the denominator is $5-5=0$. So the form is $\dfrac{-2}{0}$, not $\dfrac{0}{0}$.

2. 2

   See why the usual conjugate trick fails here. Multiplying by $\dfrac{\sqrt{x+11}+6}{\sqrt{x+11}+6}$ gives numerator $(x+11)-36=x-25$, so

   $\frac{\sqrt{x+11}-6}{x-5}=\frac{x-25}{(x-5)\left(\sqrt{x+11}+6\right)}$

   At $x=5$ the new numerator is $-20\ne 0$; the factor $(x-5)$ is still alone in the denominator. Nothing cancels, which is exactly what a genuine $\tfrac{c}{0}$ looks like. (A common slip is to "factor" $x-25$ as $(x-5)(x+5)$ — but $(x-5)(x+5)=x^2-25$, not $x-25$.)

3. 3

   Take the two one-sided limits. Near $x=5$ the numerator is close to $-2<0$, so the sign of the quotient is the opposite of the sign of $x-5$:

   $\lim_{x\to 5^{+}}\frac{\sqrt{x+11}-6}{x-5}=-\infty,\qquad \lim_{x\to 5^{-}}\frac{\sqrt{x+11}-6}{x-5}=+\infty$

4. 4

   Confirm numerically. At $x=5.0001$ the quotient is about $-2.0\times10^{4}$; at $x=4.9999$ it is about $+2.0\times10^{4}$. The two sides run off in opposite directions.

5. 5

   State the conclusion. Because the one-sided limits disagree (and are not even finite), the two-sided limit does not exist; the line $x=5$ is a vertical asymptote of this function. The general rule worth keeping: only $\tfrac00$ is indeterminate — a nonzero number over zero is always an infinite or non-existent limit, never a finite value.