Limits Plus: III

Calculus Level 2

$\lim _{ x\rightarrow 1 }{ \frac{ \sqrt [ 3 ]{ x}-1 } {\sqrt{x}-1}}=\frac{a}{b}$ ,

evaluate $a+b$ .

NOTE:

No L'Hôpital's Rule.

Problem credit: Calculus: 6E, James Stewart

The answer is 5.

2 solutions

Brian Charlesworth
Aug 29, 2014

Let $u = \sqrt[6]{x}$ . Then $u \rightarrow 1$ as $x \rightarrow 1$ , and the limit then becomes

$\lim_{u \rightarrow 1} \dfrac{u^{2} - 1}{u^{3} - 1} = \lim_{u \rightarrow 1} \dfrac{(u - 1)(u + 1)}{(u - 1)(u^{2} + u + 1)} = \lim_{u \rightarrow 1} \dfrac{u + 1}{u^2 + u + 1} = \dfrac{2}{3}$ .

Thus $a = 2, b = 3$ and $a + b = \boxed{5}$ .

A simpler approach would be to make the substitution $y=\sqrt{x}$ and then use the identity $\displaystyle \lim_{y\to a} \left( \dfrac{y^n-a^n}{y-a}\right)=na^{n-1}$

Prasun Biswas - 6 years, 4 months ago

perfect answer

Moemen Adel - 6 years, 8 months ago

Daniel Liu
Aug 29, 2014

Why no L'Hopital's rule? $\ddot\frown$

L'Hopital's rule gives $\lim\limits_{x\to 1} \dfrac{2\sqrt{x}}{3\sqrt[3]{x}^2}=\dfrac{2}{3}$ so our answer is $\boxed{5}$ .

I know right, L'Hopital makes this problem a one-liner :D.

Jayakumar Krishnan - 6 years, 9 months ago

Exactly. That's why we don't use it. The point is to demonstrate your algebraic skills.

John M. - 6 years, 9 months ago

Limits Plus: III

The answer is 5.

2 solutions

0 pending reports