MA 161 - Lesson 5: Computing Limits, Limit Laws, Squeeze Theorem

Review Example from Previous Lesson

Graph Description: The graph shows various points and discontinuities. At \(x = 3\), \(f(3)\) is not defined (shown as an open circle), but the limit exists and equals 1. At \(x = -2\), \(f(-2) = 2\) but the limit equals -1, showing that function value and limit can be different.

From the graph:

Finding Limits of Simple Functions

Constant Functions

Example: \(f(x) = 7\) (constant)
\[\lim_{x \to 197} f(x) = \lim_{x \to 197} 7 = 7\]

The limit of a constant function is always the constant, regardless of what \(x\) approaches.

Linear Functions

Example: \(f(x) = x\)
\[\lim_{x \to 197} x = 197\]
Example: \(f(x) = 7x - 5\)
\[\lim_{x \to 3} f(x) = \lim_{x \to 3} (7x - 5) = 7(3) - 5 = 16\]

Finding Limits of Polynomial Functions

Polynomial Function: \(P(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0\)

Where \(n\) is the degree of the polynomial.

Limit of Polynomial: For any polynomial \(P(x)\) and any real number \(a\):

\[\lim_{x \to a} P(x) = P(a)\]

Just substitute the value!

Example: \(P(x) = x^2 + 3x - 17\)

Find \(\lim_{x \to 2} P(x)\):

\[\lim_{x \to 2} P(x) = P(2) = 2^2 + 3(2) - 17 = 4 + 6 - 17 = -7\]

Find \(\lim_{x \to 3} P(x)\):

\[\lim_{x \to 3} P(x) = P(3) = 3^2 + 3(3) - 17 = 9 + 9 - 17 = 1\]

Breaking Down Using Limit Laws

We can also compute this step by step:

\[\lim_{x \to 2} (x^2 + 3x - 17) = \lim_{x \to 2} x^2 + \lim_{x \to 2} 3x - \lim_{x \to 2} 17\]

Evaluating each part:

Therefore: \(4 + 6 - 17 = -7\)

Limit Laws

Limit Laws: Suppose \(\lim_{x \to a} f(x) = L\) and \(\lim_{x \to a} g(x) = M\). Then:

  1. Sum Rule: \(\lim_{x \to a} [f(x) + g(x)] = L + M\)
  2. Difference Rule: \(\lim_{x \to a} [f(x) - g(x)] = L - M\)
  3. Product Rule: \(\lim_{x \to a} [f(x) \cdot g(x)] = L \cdot M\)
  4. Quotient Rule: \(\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{L}{M}\) when \(M \neq 0\)
  5. Power Rule: \(\lim_{x \to a} [f(x)]^n = L^n\)
  6. Constant Multiple Rule: \(\lim_{x \to a} [c \cdot f(x)] = c \cdot L\)

Finding Limits of Rational Functions

Rational Function: \(R(x) = \frac{P(x)}{Q(x)}\) where \(P\) and \(Q\) are polynomials.

Key Principle: If \(Q(a) \neq 0\), then \(x = a\) is in the domain of \(R(x)\) and:

\[\lim_{x \to a} R(x) = R(a) = \frac{P(a)}{Q(a)}\]
Example: \(\lim_{x \to 7} \frac{x^2 - 4x - 21}{x - 4}\)

Solution:

First, check if we can plug in \(x = 7\):

\[\lim_{x \to 7} \frac{x^2 - 4x - 21}{x - 4} = \frac{7^2 - 4(7) - 21}{7 - 4} = \frac{49 - 28 - 21}{3} = \frac{0}{3} = 0\]

When Direct Substitution Gives 0/0

Example: \(\lim_{x \to 7} \frac{x^2 - 4x - 21}{x - 7}\)

When \(x = 7\): numerator = \(49 - 28 - 21 = 0\), denominator = \(0\)

We get the indeterminate form \(\frac{0}{0}\), so we need to factor.

Solution:

Factor the numerator: \(x^2 - 4x - 21 = (x - 7)(x + 3)\)

\[\lim_{x \to 7} \frac{x^2 - 4x - 21}{x - 7} = \lim_{x \to 7} \frac{(x - 7)(x + 3)}{x - 7}\]

Cancel the common factor \((x - 7)\):

\[= \lim_{x \to 7} (x + 3) = 7 + 3 = 10\]

We can plug in because \(x + 3\) is a polynomial (and thus continuous) near \(x = 7\).

Rationalization Technique

For limits involving square roots that result in \(\frac{0}{0}\), we can use rationalization.

Rationalization Identity: \((\sqrt{a} - \sqrt{b})(\sqrt{a} + \sqrt{b}) = a - b\)

Example: \(\lim_{x \to 0} \frac{\sqrt{x + 5} - \sqrt{5}}{x}\)

Direct substitution gives \(\frac{0}{0}\), so we rationalize:

Solution:

\[\lim_{x \to 0} \frac{\sqrt{x + 5} - \sqrt{5}}{x} \cdot \frac{\sqrt{x + 5} + \sqrt{5}}{\sqrt{x + 5} + \sqrt{5}}\] \[= \lim_{x \to 0} \frac{(x + 5) - 5}{x(\sqrt{x + 5} + \sqrt{5})}\] \[= \lim_{x \to 0} \frac{x}{x(\sqrt{x + 5} + \sqrt{5})}\] \[= \lim_{x \to 0} \frac{1}{\sqrt{x + 5} + \sqrt{5}}\] \[= \frac{1}{\sqrt{0 + 5} + \sqrt{5}} = \frac{1}{2\sqrt{5}}\]

Squeeze (Sandwich) Theorem

Squeeze Theorem: Suppose \(f(x) \leq g(x) \leq h(x)\) near \(x = a\).

If \(\lim_{x \to a} f(x) = \lim_{x \to a} h(x) = L\), then \(\lim_{x \to a} g(x) = L\)

Graph Description: The graph shows three functions where \(g(x)\) is "squeezed" between \(f(x)\) and \(h(x)\). Near \(x = a\), both \(f(x)\) and \(h(x)\) approach the same limit \(L\), forcing \(g(x)\) to also approach \(L\).

Example: \(\lim_{x \to 0} x^2 \cos\left(\frac{1}{x}\right)\)

Solution:

We know that for any \(x \neq 0\): \(-1 \leq \cos\left(\frac{1}{x}\right) \leq 1\)

Multiply all parts by \(x^2\) (which is always non-negative):

\[-x^2 \leq x^2 \cos\left(\frac{1}{x}\right) \leq x^2\]

Now find the limits of the "squeeze" functions:

By the Squeeze Theorem:

\[\lim_{x \to 0} x^2 \cos\left(\frac{1}{x}\right) = 0\]

Key Insight: The Squeeze Theorem is particularly useful when dealing with oscillating functions (like trigonometric functions) that are bounded by simpler functions.