# Find $\lim_{x \to 0 }\frac{(1+x)^{(1/2)} -1}{(1+x)^{(1/3)} -1}$

Find the limit without using L'hopital

$$\lim_{x \to 0 }\frac{(1+x)^{(1/2)} -1}{(1+x)^{(1/3)} -1}$$

I have tried multiply by the conjugate of the denominator and numerator but didn't work

Any hint would be appreciated

Thank you

• You should try the formula $$\lim_{x\to a} \frac{x^n-a^n} {x-a} =na^{n-1}$$ – Paramanand Singh Aug 20 '18 at 5:34

There are two conjugates you need to multiply:

\begin{align*} \lim_{x\to0} \frac{\sqrt{1 + x} - 1}{\sqrt[3]{1 + x} - 1} &= \lim_{x\to0} \frac{\sqrt{1 + x} - 1}{\sqrt[3]{1 + x} - 1} \cdot \frac{\sqrt{1 + x} + 1}{\sqrt{1 + x} + 1} \cdot \frac{\sqrt[3]{1 + x}^2 + \sqrt[3]{1 + x} + 1}{\sqrt[3]{1 + x}^2 + \sqrt[3]{1 + x} + 1} \\ &= \lim_{x\to0} \frac{(\sqrt{1 + x} - 1)(\sqrt{1 + x} + 1)}{(\sqrt[3]{1 + x} - 1)(\sqrt[3]{1 + x}^2 + \sqrt[3]{1 + x} + 1)} \cdot \frac{\sqrt[3]{1 + x}^2 + \sqrt[3]{1 + x} + 1}{\sqrt{1 + x} + 1} \\ &= \lim_{x\to0} \frac{\sqrt{1 + x}^2 - 1^2}{\sqrt[3]{1 + x}^3 - 1^3} \cdot \frac{\sqrt[3]{1 + x}^2 + \sqrt[3]{1 + x} + 1}{\sqrt{1 + x} + 1} \\ &= \lim_{x\to0} \frac{\sqrt[3]{1 + x}^2 + \sqrt[3]{1 + x} + 1}{\sqrt{1 + x} + 1} \\ &= \frac{\sqrt[3]{1 + 0}^2 + \sqrt[3]{1 + 0} + 1}{\sqrt{1 + 0} + 1} = \frac{3}{2}. \end{align*}

Let $1+x=(1+y)^6$.

The expression becomes

$\dfrac{(1+y)^3-1}{(1+y)^2-1} =\dfrac{3y+3y^2+y^3}{2y+y^2} =\dfrac{3+3y+y^2}{2+y} \to \dfrac32$.

Generalizations should be clear.

You can use the fact that for $x$ near $0$, $$(1+x)^\alpha = 1+\alpha x+o(x)$$ Then, in the limit $x\to0$, \begin{align} \lim_{x\to0} \frac{(1+x)^{1/2}-1}{(1+x)^{1/3}-1} &= \lim_{x\to 0}\frac{1+x/2+o(x)-1} {1+x/3+o(x)-1} \\ &=\lim_{x\to0} \frac{x/2+o(x)}{x/3+o(x)} \\ &= \lim_{x\to 0} \frac{3}{2} \\ &=\frac{3}{2} \end{align}

• It is better to use little o notation like $(1+x)^n=1+nx+o(x)$ rather than replacing $(1+x)^n$ by $1+nx$. – Paramanand Singh Aug 20 '18 at 5:31
• Ok thank you for the comment – Zachary Aug 20 '18 at 5:35

Using that $t^2-1=(t-1)(t+1)$ and $t^3-1=(t-1)(t^2+t+1)$ we have

$$\lim_{x \to 0}\dfrac{\sqrt{x+1}-1}{\sqrt[3]{x+1} -1}=\lim_{x \to 0}\left[\dfrac{\sqrt{x+1}-1}{\sqrt[3]{x+1} -1}\cdot \dfrac{\sqrt{x+1}+1}{\sqrt{x+1}+1}\cdot \dfrac{\sqrt[3]{(x+1)^2}+\sqrt[3]{x+1}+1}{\sqrt[3]{(x+1)^2}+\sqrt[3]{x+1}+1}\right]$$ and thus

$$\lim_{x \to 0}\dfrac{\sqrt{x+1}-1}{\sqrt[3]{x+1} -1}=\lim_{x \to 0}\left[\dfrac{\sqrt[3]{(x+1)^2}+\sqrt[3]{x+1}+1}{\sqrt{x+1}+1}\right]=\dfrac 32.$$

Hint:

As lcm$(2,3)=6$

choose $\sqrt[6]{1+x}=y\implies\sqrt[2]{1+x}=y^3,\sqrt[3]{1+x}=y^2$

and as $x\to0,y\to1$

OR

choose $\sqrt[6]{1+x}=z+1\implies z\to0$

Let $y:=1+x;$ $y>0$.

$\dfrac{y^{1/2}-1}{y^{1/3}-1}=$

$\dfrac{y -1}{y^{1/2}+1}\dfrac{y^{2/3}+y^{1/3}+1}{y-1}.$

The limit $y \rightarrow 1$ is?

Used:

$y-1=(y^{1/2}-1)(y^{1/2}+1)$, and

$y-1=(y^{1/3}-1)(y^{2/3}+y^{1/3} +1)$.