0
$\begingroup$

How can we solve this limit without using L'Hospital rule? I tried using some other methods but can't get the answer. $$\lim_{x\to 1}\frac{\sqrt[3]{x}-1}{\sqrt[]{x}-1}$$

3 Answers 3

9

Let $x=t^6$. Note that $$\lim_{x \to 1 }\frac{\sqrt[3]{x}-1}{\sqrt[]{x}-1}=\lim_{t \to 1} \frac{t^2-1}{t^3-1}=\lim_{t \to 1}\frac{t+1}{t^2+t+1}$$ So the limit is $\frac{2}{3}$.

  • 0
    How do we figure out what the variable substitution should be? I was trying $$t = \sqrt[3]{x}$$ and $$t =\sqrt[]{x}$$ but they left the limit in indeterminate form. – 2017-02-17
  • 1
    It has to be LCM$(2,3)$ – 2017-02-17
  • 0
    @lakada We want the numerator and denominator in forms we can factorize. – 2017-02-17
0

Note:

$$\frac{\sqrt[3]x-1}{\sqrt x-1}=\frac{\sqrt[3]x-\sqrt[3]1}{x-1}\left(\frac{\sqrt x-\sqrt1}{x-1}\right)^{-1}$$

Thus, the limit is the derivative of some functions. Let $f(x)=\sqrt[3]x$ and $g(x)=\sqrt x$ so that we have

$$\lim_{x\to1}\frac{\sqrt[3]x-1}{\sqrt x-1}=\frac{f'(1)}{g'(1)}=\frac23$$

0

Divide numerator and denominator by $x-1$ and use the standard limit $$\lim_{x\to a} \frac{x^{n} - a^{n}} {x-a} =na^{n-1}$$ to get the answer as $(1/3)/(1/2)=2/3$.