Limits from Graphs and Algebra

Develop approaching a value, removable discontinuities.

Lesson 1 of 32 parts Practice Flashcards

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Approaching a Value

Objective: Explain approaching a value, apply the rule, and justify relevant restrictions.

A limit L describes the output approached as x approaches a, independently of whether f(a) is defined or equals L. For a two-sided finite limit, the left and right limits must agree. Polynomial functions are continuous, so substitution evaluates their limits. A plotted point at x=a specifies a function value, while the nearby curve determines the limit.

Formulas

For a polynomial p: limₓ→a p(x)=p(a)

Worked examples

Try itFind limₓ→7 (3x²+5).Show answer

Step 1: A polynomial is continuous at every real input. Step 2: Substitute x=7: 3(7)²+5. Step 3: The limit is 152.

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Approaching a Value

Objective: Explain approaching a value, apply the rule, and justify relevant restrictions.

Formulas

For a polynomial p: limₓ→a p(x)=p(a)

Worked examples

Try itFind limₓ→7 (3x²+5).Show answer

Step 1: A polynomial is continuous at every real input. Step 2: Substitute x=7: 3(7)²+5. Step 3: The limit is 152.

Removable Discontinuities

Objective: Explain removable discontinuities, apply the rule, and justify relevant restrictions.

A quotient that produces 0/0 under substitution has an indeterminate form, not a limit value. Factoring and cancellation can produce an equivalent rule on a punctured neighborhood of the target point. Evaluate the limit using that rule, retaining the fact that the original denominator vanishes at the point. The limit may exist even if the original function has a hole.

Formulas

$(x²−a²)/(x−a)=x+a for x≠a$

Worked examples

Try itFind limₓ→7 (x²−49)/(x−7).Show answer

Step 1: Factor x²−49=(x−7)(x+7). Step 2: For x≠7, the quotient equals x+7. Step 3: The limit is 7+7=14, even though the original function is undefined there.

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