Mathfolis

Algebraic Limit Evaluation

Unit 1 · Limits & Continuity

Evaluating Limits Algebraically

When direct substitution gives 0/0 (indeterminate form), use algebraic manipulation: factor and cancel, rationalize with conjugates, or simplify complex fractions. If bounds can be found on both sides, the Squeeze Theorem gives the limit directly.

Strategy

Step 1 — Direct substitution: if defined, done.
Step 2 — If indeterminate
00⇒factor, rationalize, or simplify\tfrac{0}{0} \Rightarrow \text{factor, rationalize, or simplify}
Step 3 — Substitute after simplification.

Special Limits

lim⁡x→0sin⁡xx=1lim⁡x→01−cos⁡xx=0\lim_{x \to 0} \frac{\sin x}{x} = 1 \qquad \lim_{x \to 0} \frac{1 - \cos x}{x} = 0

Squeeze Theorem

If g(x) ≤ f(x) ≤ h(x) near c and lim g = lim h = L, then
lim⁡x→cf(x)=L\lim_{x \to c} f(x) = L
Caution: Tables and graphs can suggest a limit but don't prove it. Always use algebraic methods for exact values.
Type 1

Factoring to Remove Indeterminate Form

Factor the numerator (or denominator), cancel the common zero factor, then substitute.

Example 1
Evaluate the limit.
lim⁡x→−3x2+x−6x+3\lim_{x \to -3} \frac{x^2 + x - 6}{x + 3}

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Type 2

Rationalization

When a square root creates the indeterminate form, multiply by the conjugate to clear the radical from the numerator or denominator.

Example 2
Evaluate the limit.
lim⁡x→9x−3x−9\lim_{x \to 9} \frac{\sqrt{x} - 3}{x - 9}

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Type 3

Squeeze Theorem

Bound the oscillating function between two functions that share the same limit. The target function is squeezed to the same value.

Example 3
Evaluate the limit using the Squeeze Theorem.
lim⁡x→0x4cos⁡ ⁣(1x)\lim_{x \to 0} x^4 \cos\!\left(\frac{1}{x}\right)

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