Limits Notes: Conjugates, Piecewise, and Infinity

Conjugate trick for square-root limits - Target limit (part a):

$\lim_{x\to 36} \frac{\sqrt{x}-6}{x-36}$

• When you plug in $x=36$, you get $0/0$, which means we need to simplify the expression.

• To simplify, multiply the top and bottom by the "conjugate" of the square-root part, which is $\sqrt{x}+6$:
  $\frac{\sqrt{x}-6}{x-36}\cdot\frac{\sqrt{x}+6}{\sqrt{x}+6} = \frac{(\sqrt{x}-6)(\sqrt{x}+6)}{(x-36)(\sqrt{x}+6)}$

• Use the difference of squares rule ($(a-b)(a+b) = a^2-b^2$) on the top part. Here, $a=\sqrt{x}$ and $b=6$, so $(\sqrt{x}-6)(\sqrt{x}+6) = x-36$.

• This simplifies the expression to:
  $\frac{x-36}{(x-36)(\sqrt{x}+6)} = \frac{1}{\sqrt{x}+6},\quad x\neq 36.$

• Now that the problem factor $(x-36)$ is gone, plug in $x=36$ again:
  $\frac{1}{\sqrt{36}+6} = \frac{1}{6+6} = \frac{1}{12}.$

• Key takeaway: If you have a square root expression (like $\sqrt{x}-6$) in the numerator and direct substitution gives $0/0$, multiply by its conjugate to help simplify and cancel out the problematic terms.

• Target limit (part b):
  $\lim_{h\to 0} \frac{\frac{6}{5+h}-\frac{6}{5}}{h}$

• If you plug in $h=0$, you also get $0/0$. So, we need to simplify.

• First, combine the two fractions in the numerator into a single fraction. Find a common denominator, which is $5(5+h)$.
  $<br>\frac{6}{5+h}-\frac{6}{5} = \frac{6\cdot 5}{5(5+h)} - \frac{6(5+h)}{5(5+h)} = \frac{30 - (30+6h)}{5(5+h)} = \frac{30 - 30 - 6h}{5(5+h)} = \frac{-6h}{5(5+h)}.<br>$

• Now, substitute this back into the original expression:
  $\frac{\frac{-6h}{5(5+h)}}{h} = \frac{-6h}{h\cdot 5(5+h)}.$

• The $h$ in the top and bottom cancels out (since $h\neq 0$ in the limit approaching 0):
  $\frac{-6}{5(5+h)}.$

• Finally, plug in $h=0$ to find the limit:
  $\lim_{h\to 0} \frac{-6}{5(5+h)} = \frac{-6}{5(5+0)} = \frac{-6}{25}.$

• Important note: Be careful when you subtract expressions, especially when a minus sign is involved, like $-(30+6h) = -30-6h$. Make sure the minus sign applies to all terms inside the parentheses.

Piecewise functions: one-sided limits at boundary points

• A piecewise function changes its rule (formula) at certain points. We need to check limits at these "boundary points" to see what the function is approaching.
  
  f(x) = \begin{cases}
  x+1, & x \le 2, \
  2x-1, & 2 4.
  \end{cases}
  

• Here, the rules change at $x=2$ and $x=4$.

1. Boundary at $x=2$:

   • Left-hand limit: As $x$ approaches 2 from values less than 2 (e.g., 1.9, 1.99), use the rule $x+1$:
     $\lim_{x\to 2^-} f(x) = 2+1 = 3.$

   • Right-hand limit: As $x$ approaches 2 from values greater than 2 (e.g., 2.1, 2.01), use the rule $2x-1$:
     $\lim_{x\to 2^+} f(x) = 2\cdot 2 - 1 = 3.$

   • Since both the left-hand and right-hand limits are the same (both are 3), the limit at $x=2$ exists:
     $\lim_{x\to 2} f(x) = 3.$

2. Boundary at $x=4$:

   • Left-hand limit: As $x$ approaches 4 from values less than 4 (e.g., 3.9, 3.99), use the rule $2x-1$:
     $\lim_{x\to 4^-} f(x) = 2\cdot 4 - 1 = 7.$

   • Right-hand limit: As $x$ approaches 4 from values greater than 4 (e.g., 4.1, 4.01), use the rule $10x$:
     $\lim_{x\to 4^+} f(x) = 10\cdot 4 = 40.$

   • Since the left-hand limit (7) and the right-hand limit (40) are not the same, the two-sided limit at $x=4$ does not exist:
     $\lim_{x\to 4} f(x) \text{ does not exist}.$

3. Conceptual takeaway:

   • A limit for a function at a point only exists if the value it approaches from the left is the same as the value it approaches from the right.

   • When a function changes its definition (rule) at a point, you must check both the left-hand and right-hand limits separately to determine if the overall limit exists.

4. Quick follow-up example (using given limits of $f$ and $g$ without knowing their formulas):

   • Given that when $x$ approaches 4, $f(x)$ approaches 16, and $g(x)$ approaches 9:
     $\lim<em>{x\to 4} f(x) = 16\quad\text{and}\quad \lim</em>{x\to 4} g(x) = 9,$

   • a) The limit of a square root is the square root of the limit:
     $\lim<em>{x\to 4} \sqrt{f(x)} = \sqrt{\lim</em>{x\to 4} f(x)} = \sqrt{16} = 4.$

   • b) The limit of a sum/difference is the sum/difference of the limits, and constants can be pulled out:
     $\lim<em>{x\to 4} (3f(x) - g(x)) = 3\cdot (\lim</em>{x\to 4} f(x)) - (\lim_{x\to 4} g(x)) = 3\cdot 16 - 9 = 48 - 9 = 39.$

   • Note: We don't need to know the actual formulas for $f(x)$ and $g(x)$; we only use the given values of their limits and standard limit rules.