1
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ You should try the formula $$\lim_{x\to a} \frac{x^n-a^n} {x-a} =na^{n-1}$$ $\endgroup$ – Paramanand Singh Aug 20 '18 at 5:34
4
$\begingroup$

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*}

$\endgroup$
4
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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}

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

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

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

$\endgroup$
1
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.