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How to Solve Limits

A limit asks what a function value approaches as x gets close to a number. Most limits are quick: you try substitution, and if that fails with a 0/0 form, you do one clean algebra move (factor/cancel or rationalize) and try again.

∞ Calculus⏱️ ~23 min read🔴 Advanced

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The One Sentence Definition

You do not have to hit x = a exactly. The statement lim(x→a) f(x) = L means: when x gets close to a (from either side), f(x) gets close to L.

lim(x→a) f(x) = L  (approach, not equal)

That’s why limits can exist even if the function is undefined at a point (like a hole after cancellation). Limits are about behavior near a point.

A Step-by-Step Workflow

  1. Step 1: Substitute x = a.
    • If you get a real number: you’re done.
    • If you get 0/0: you need algebra.
    • If you get something like 5/0: check one-sided or infinite behavior.
  2. Step 2: For 0/0, do one of the standard fixes: factoring/canceling, common denominator, or rationalizing.
  3. Step 3: Substitute again after simplifying.
  4. Step 4: If the problem involves absolute value, piecewise, or roots, compute left and right limits.
  5. Step 5: Conclude: limit exists and equals a number, or limit is ±∞, or DNE.

If you want a clean overview of the common limit techniques and what they mean graphically, OpenStax gives a friendly introduction in its section on limits of functions. The big win is learning the small set of patterns that appear over and over.

Example 1: Direct Substitution Works

Problem: Evaluate lim(x→3) (x^2 + 2x − 1).

Polynomials are continuous everywhere, so substitution is valid:

lim(x→3) (x^2 + 2x − 1) = 3^2 + 2·3 − 1 = 9 + 6 − 1 = 14

Example 2: Factor and Cancel (0/0)

Problem: Evaluate lim(x→1) (x^2 − 1)/(x − 1).

Substituting gives 0/0, so factor:

(x^2 − 1) = (x − 1)(x + 1)
(x^2 − 1)/(x − 1) = (x − 1)(x + 1)/(x − 1) = x + 1  (for x ≠ 1)

Now substitute into the simplified expression:

lim(x→1) (x + 1) = 2

Interpretation: the original function has a hole at x = 1, but the values approach 2.

Example 3: Rationalize (Square Roots)

Problem: Evaluate lim(x→0) (√(x + 9) − 3)/x.

Substitution gives (3 − 3)/0 = 0/0. Rationalize by multiplying by the conjugate:

(√(x+9) − 3)/x · (√(x+9) + 3)/(√(x+9) + 3)
= ((x+9) − 9) / (x(√(x+9) + 3))
= x / (x(√(x+9) + 3))
= 1 / (√(x+9) + 3)

Now substitute x = 0:

1/(√(9) + 3) = 1/6

When You Must Use One-Sided Limits

A two-sided limit exists only if the left-hand limit and right-hand limit match:

lim(x→a) f(x) exists ⇔ lim(x→a−) f(x) = lim(x→a+) f(x)

Typical “one-sided” triggers: absolute value, piecewise functions, and rational functions with vertical asymptotes.

Example 4: A Limit That Does Not Exist

Problem: Evaluate lim(x→0) |x|/x.

For x > 0, |x| = x, so |x|/x = 1. For x < 0, |x| = −x, so |x|/x = −1.

lim(x→0+) |x|/x = 1
lim(x→0−) |x|/x = −1

Since the one-sided limits are different, the two-sided limit does not exist.

Infinite Limits and Vertical Asymptotes

Sometimes substitution produces division by zero that does not cancel (like 5/0). In that case, the function may blow up to +∞ or −∞. The sign depends on the direction you approach and the signs of numerator/denominator.

Fast sign check:

  • Pick a number slightly less than a (for left-hand) and slightly more than a (for right-hand).
  • Evaluate the sign (positive/negative), not the exact value.
  • Conclude whether it heads to +∞ or −∞ on each side.

Special Trig Limits (The Ones Worth Memorizing)

The most important trig limit is:

lim(x→0) sin(x)/x = 1

Many trig limits are solved by rewriting the expression until you can use that pattern. For example:

Example: Evaluate lim(x→0) (sin(5x))/(x).

Multiply and divide by 5 to create sin(5x)/(5x):

sin(5x)/x = 5 · (sin(5x)/(5x))
lim(x→0) sin(5x)/x = 5 · lim(x→0) sin(5x)/(5x) = 5 · 1 = 5

The key idea is “make the denominator match what’s inside sine.”

Limits at Infinity (Rational Functions)

Another common category is what happens as x → ∞ or x → −∞. For rational functions (polynomial over polynomial), the limit is controlled by the highest powers.

Example: Evaluate lim(x→∞) (2x^2 − x + 1)/(x^2 + 4).

Divide top and bottom by x^2:

(2x^2 − x + 1)/(x^2 + 4)
= (2 − 1/x + 1/x^2) / (1 + 4/x^2)
lim(x→∞) = (2 − 0 + 0)/(1 + 0) = 2

Shortcut: if the degrees match, the limit is the ratio of leading coefficients.

Common Mistakes

Stopping at 0/0

0/0 means “try again after simplifying,” not “the limit is 0.” Factor/cancel or rationalize.

Canceling incorrectly

You can only cancel factors, not terms. For example, you can cancel (x − 1) if it is a factor, not if it’s part of a sum.

Forgetting one-sided limits

For absolute values and piecewise functions, you must check left and right. Matching is required for the limit to exist.

Confusing DNE with ∞

If both sides head to +∞, the limit is +∞ (an infinite limit). If one side is +∞ and the other is −∞, the two-sided limit DNE.

Practice Problems

  1. lim(x→2) (x^3 − 8)/(x − 2)
  2. lim(x→0) (√(1 + x) − 1)/x
  3. lim(x→0) |x|/x (state whether it exists)
  4. lim(x→1) (x − 1)/|x − 1| (check one-sided)

If you get 0/0, write the simplification move you used (factor, conjugate, or common denominator). That is usually what the teacher is grading.

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How to Solve Limits | Step-by-Step Methods + Examples