Calculus & Vectors (MCV4U) · Rate of Change

Evaluating limits

Evaluate limits by direct substitution and by factoring to remove a 0/0 indeterminate form. Unlimited questions, five difficulty levels, and a full worked solution every time — free.

Practice this skill Curriculum: MCV4U-RC2.1, MCV4U-RC3.1

Try one

Evaluate the limit.

lim x->2 −3x2 + 3

Answer: -9

See one solved, step by step

Evaluate the limit.

lim x->1 (x2 - 4x + 3) / (x - 1)
📘 Worked solution
1Substituting x = 1 gives 00Both the top and bottom equal 0 at x = 1, so this is an indeterminate form — we must simplify first.
2x2 - 4x + 3 = (x - (1))(x - 3)Factor the numerator; x = 1 and x = 3 are its roots.
3(x - (1))(x - 3) / (x - 1) = x - 3The (x - (1)) factor cancels top and bottom, valid since x is only approaching 1, never equal to it.
4lim x->1 x - 3 = −2Now this is continuous at x = 1 — substitute directly.

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