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Wize High School Grade 12 Calculus Textbook > Rate of Change

Evaluating 00\frac{0}{0} Limits -- with Substitution

Evaluating 00\frac{0}{0} Limits -- with Absolute ValuesPrevious Section
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Example: Substitution 0/0 Limits

Evaluate the limit lim⁡x→64 x3−4x−8\displaystyle \lim_{x\to64}\ \frac{\sqrt[3]{x}-4}{\sqrt{x}-8}.

We can't factor the numerator and denominator the way it currently is, but we notice that there are radicals in the numerator and denominator.

The xx term in the numerator is x3=x1/3\sqrt[3]x=x^{1/3}, the xx term in the denominator is x=x1/2\sqrt x=x^{1/2} → the LCD is x1/6x^{1/6}.

So, we let u=x1/6u=x^{1/6}, then
  • x3=x1/3=(x1/6)2 → x3=u2\sqrt[3]x=x^{1/3}=\left(x^{1/6}\right)^2\ \to \ \boxed{\sqrt[3]x=u^2}
  • x=x1/2=(x1/6)3 → x=u3\sqrt x=x^{1/2}=\left(x^{1/6}\right)^3\ \to\ \boxed{\sqrt x=u^3}
  • As x→64x\to 64, u→(64)1/6=2\boxed{u\to(64)^{1/6}=2}
Substituting this back into our limit expression:
lim⁡x→64 x3−4x−8\displaystyle \lim_{x\to64}\ \frac{\sqrt[3]{x}-4}{\sqrt{x}-8}
=lim⁡u→2 u2−4u3−8=\displaystyle \lim_{u\to2}\ \frac{u^2-4}{u^3-8}

Recall that a2−b2=(a−b)(a+b)\pink{a^2-b^2=\left(a-b\right)\left(a+b\right)} and a3−b3=(a−b)(a2+ab+b2)\pink{a^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)}

=lim⁡u→2 (u−2)(u+2)(u−2)(u2+2u+4)\displaystyle =\lim_{u\to2}\ \frac{\left(u-2\right)\left(u+2\right)}{\left(u-2\right)\left(u^2+2u+4\right)}
=lim⁡u→2 u+2u2+2u+4\displaystyle =\lim_{u\to2}\ \frac{u+2}{u^2+2u+4}
=2+222+2(2)+4\displaystyle =\frac{2+2}{2^2+2\left(2\right)+4}
=44+4+4\displaystyle =\frac{4}{4+4+4}
=13\displaystyle =\frac{1}{3}
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Example: Substitution 0/0 Limits

Evaluate lim⁡x→0 (x+27)13−3x\displaystyle \lim_{x\to0}\ \frac{\left(x+27\right)^{\frac{1}{3}}-3}{x}.

We can't factor the numerator and denominator the way it currently is, but we notice that there are radicals in the numerator and denominator.

The xx term in the numerator is 1/3{1/3}, the xx term in the denominator is 1/1{1/1} → the LCD is 1/3{1/3}.

So, we let u=(x+27)1/3u=(x+27)^{1/3}, then
  • (x+27)1/3=u\boxed{(x+27)^{1/3}=u}
  • u3=x+27 → x=u3−27u^3=x+27\ \to\ \boxed{x=u^3-27}
  • As x→0x\to0, u→(0+27)1/3=3\boxed{u\to(0+27)^{1/3}=3}
Substituting this back into our limit expression:
lim⁡x→0 (x+27)13−3x\displaystyle \lim_{x\to0}\ \frac{\left(x+27\right)^{\frac{1}{3}}-3}{x}
=lim⁡u→3 u−3u3−27=\displaystyle \lim_{u\to3}\ \frac{u-3}{u^3-27}

Recall that a3−b3=(a−b)(a2+ab+b2)\pink{a^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)}

=lim⁡u→3 u−3(u−3)(u2+3u+9)\displaystyle =\lim_{u\to3}\ \frac{u-3}{\left(u-3\right)\left(u^2+3u+9\right)}
=lim⁡u→3 1u2+3u+9\displaystyle =\lim_{u\to3}\ \frac{1}{u^2+3u+9}
=1(3)2−3(3)+9\displaystyle =\frac{1}{(3)^2-3(3)+9}
=19+9+9\displaystyle =\frac{1}{9+9+9}
=127\displaystyle =\frac{1}{27}

Practice: Substitution 0/0 Limits

Evaluate lim⁡x→0 x+16−1x\displaystyle \lim_{x\to0}\ \frac{\sqrt[6]{x+1}-1}{\sqrt{x}}.