Try each one on paper first, then check a single step. That is worth far more than reading a finished solution.
Example 1 — removing an indeterminate form
Evaluate x→3limx2−x−6x2−9.
- 1Substitute first: 9−3−69−9=00Indeterminate — the expression must be simplified.
- 2Factorise the numerator: x2−9=(x−3)(x+3)Difference of two squares.
- 3Factorise the denominator: x2−x−6=(x−3)(x+2)The (x−3) appearing in both is what caused the 00.
- 4Cancel the common factor, valid because x=3 in a limit.The limit deliberately excludes the point itself.
- 5limx→3x+2x+3=56Now substitution works.
Example 2 — a trigonometric limit
Evaluate x→0limsin2xsin5x.
- 1Substituting gives 00, so use limθ→0θsinθ=1.Radians throughout.
- 2The rule needs a matching denominator, so manufacture one for each part.
- 3sin2xsin5x=2xsin2x⋅2x5xsin5x⋅5xMultiplying and dividing by 5x and 2x changes nothing.
- 4Both fractions tend to 1, leaving 2x5x.
- 5=25For small angles, sinkx behaves like kx.
Example 3 — differentiating from first principles
Differentiate f(x)=x2 from first principles.
- 1f′(x)=limh→0hf(x+h)−f(x)The gradient of a chord, as its width shrinks.
- 2=limh→0h(x+h)2−x2
- 3=limh→0hx2+2xh+h2−x2Expand before doing anything else.
- 4=limh→0h2xh+h2=limh→0(2x+h)Cancel h only after simplifying — h=0 inside the limit.
- 5=2xKeep the limit notation on every line until this final substitution.