Limit with non square roots without l'Hôpital’s rule

Question

I want to calculate the following limit:

$\lim_{x\to0}\frac{\sqrt[3]{1+x^2}-\sqrt[4]{1-2x}}{x+x^3}$ without the l'Hôpital rule.

This limit can be found in Kitchen's book Calculus of one Variable, exercise 5t), section 3-5, during that section the book has substitution techniques and rationalization, but the latter is only done when both roots are square roots, so if there is a specific factorization that works, I also would thank how to find it.

By now i have tried the substitution $u=1+x^2$ and $u=1-x$, but none has got me anywhere.

Answer 1

Outline of Strategy: Write $\sqrt[3]{1+x^2} - \sqrt[4]{1-2x}= \sqrt[3]{1+x^2} - 1 + 1 - \sqrt[4]{1-2x}= \dfrac{x^2}{....}+ \dfrac{2x}{...}= x\left(\dfrac{1}{...}+\dfrac{2x}{...}\right)$, and for the denominator: $x+x^3 = x(1+x^2)$. Then the $x$ got canceled out and you can plug $x = 0$ into the expression to get the limit. You no longer have the indeterminate form. Note that the first part uses $a - b = \dfrac{a^3 - b^3}{a^2+ab+b^2}$,and the second part uses $a - b = \dfrac{a^4 - b^4}{a^3 + a^2b+ ab^2 + b^3}$.

Answer 2

If you are bored, you may use the analogous identity for the $12$th power:

$$\begin{align*} a^{12} - b^{12} &= (a-b)(a^{11} + a^{10}b + a^9b^2 + \cdots + b^{11})\\ &= (a-b)\sum_{i=0}^{11} a^{11-i}b^i \end{align*}$$

Let $a = \sqrt[3]{1+x^2}$ and $b = \sqrt[4]{1-2x}$. Then the required limit is

$$\begin{align*} \lim_{x\to 0}\frac{\sqrt[3]{1+x^2} - \sqrt[4]{1-2x}}{x+x^3} &= \lim_{x\to 0}\frac{a^{12}-b^{12}}{(x+x^3) \sum_{i=0}^{11} a^{11-i}b^i}\\ &= \lim_{x\to 0}\frac{(1+x^2)^{4}-(1-2x)^{3}}{(x+x^3) \sum_{i=0}^{11} a^{11-i}b^i}\\ &= \lim_{x\to 0}\frac{1+4x^2+6x^4+4x^6+x^8 - 1+6x-12x^2 +8x^3}{(x+x^3) \sum_{i=0}^{11} a^{11-i}b^i}\\ &= \lim_{x\to 0}\frac{x(4x+6x^3+4x^5+x^7 +6-12x +8x^2)}{x(1+x) \sum_{i=0}^{11} a^{11-i}b^i}\\ &= \frac{6}{12} = \frac 12 \end{align*}$$

because $\lim_{x\to0}\sum_{i=0}^{11} a^{11-i}b^i = 12$.

Answer 3

Observe that, you have :

$$ \begin{align}\lim_{x\to0}\frac{\sqrt[3]{1+x^2}-\sqrt[4]{1-2x}}{x(x^2+1)}&=\lim_{x\to0}\frac {1-\sqrt[4]{1-2x}}{x}\end{align} $$

Let $\thinspace 1-2x=u^4$ as $u\to 1$, then you obtain :

$$ \begin{align}\lim_{u\to 1}\frac {2(1-u)}{1-u^4}&=\lim_{u\to 1}\frac {2(1-u)}{(1-u)(1+u)(1+u^2)}\\ &=\frac 12\thinspace .\end{align} $$

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We used the following facts :

$$ \begin{align}\lim_{x\to0}\frac{\sqrt[3]{1+x^2}-\sqrt[4]{1-2x}}{x(x^2+1)}&=\lim_{x\to0}\frac {\sqrt[3]{1+x^2}-\sqrt[4]{1-2x}}{x}\end{align} $$

and

$$ \begin{align}\lim_{x\to 0}\frac {\sqrt [3]{1+x^2}-1}{x}&=\lim_{u\to 1}\sqrt {\frac {(u-1)^2}{u^3-1}}\\ &=0\thinspace .\end{align} $$