without L'Hospital: $\lim _{x\to 0}\frac{\sqrt[5]{1+3x^4}-\sqrt{1-2x}}{\sqrt[3]{1+x}-\sqrt{1+x}}$

Question

$$ \lim _{x\to 0}\left(\frac{\sqrt[5]{1+3x^4}-\sqrt{1-2x}}{\sqrt[3]{1+x}-\sqrt{1+x}}\right) $$

Answer 1

$$\lim _{x\to 0}\frac{\sqrt[5]{1+3x^4}-\sqrt{1-2x}}{\sqrt[3]{1+x}-\sqrt{1+x}}=$$

$$=\lim_{x\to 0}\frac{\frac{\sqrt[5]{1+3x^4}-1}{x}-\frac{\sqrt{1-2x}-1}{x}}{\frac{\sqrt[3]{1+x}-1}{x}-\frac{\sqrt{1+x}-1}{x}}$$

Use the formula/identity $$a^k-b^k=(a-b)(a^{k-1}+a^{k-2}b+\cdots+b^{k-1}).$$

$k\ge 2$, $k\in\mathbb Z$, $a,b\in\mathbb R$.

I'll find one of the limits:

$$\lim_{x\to 0}\frac{\sqrt[5]{1+3x^4}-1}{x}=$$

$$=\lim_{x\to 0}\frac{(\sqrt[5]{1+3x^4})^5-1^5}{x((\sqrt[5]{1+3x^4})^4+(\sqrt[5]{1+3x^4})^3+\cdots+1)}=$$

$$=\lim_{x\to 0}\frac{3x^3}{(\sqrt[5]{1+3x^4})^4+(\sqrt[5]{1+3x^4})^3+\cdots+1}=$$

$$=\frac{3\cdot 0^3}{(\sqrt[5]{1+3\cdot 0^4})^4+(\sqrt[5]{1+3\cdot 0^4})^3+\cdots+1}=0$$

WolframAlpha says that $-6$ is the answer of your limit.

Answer 2

By using a Taylor series expansion, one obtains, as $x \to 0$, $$ \left(1+ax^p\right)^\alpha=1+a\alpha x^p+o(x^p) $$ giving, as $x \to 0$, $$\frac{\sqrt[5]{1+3x^4}-\sqrt{1-2x}}{\sqrt[3]{1+x}-\sqrt{1+x}}= \frac{\left({1-\frac{3}{5}x^4+o(x^4)}\right)-\left({1+\frac{2}{2}x+o(x)}\right)}{{\left({1+\frac{1}{3}x+o(x)}\right)-\left({1+\frac{1}{2}x+o(x)}\right)}}=\frac{x+o(x)}{-\frac16x+o(x)}\to -6.$$

Answer 3

Hint :

Rewrite the expression as $$\frac{\sqrt[5]{1+3x^4}-\sqrt{1-2^{}x}}{\sqrt[3]{1+x}-\sqrt{1+x}}=\frac{\dfrac{\sqrt[5]{1+3x^4}-1}{x}-\dfrac{\sqrt{1-2^{}x}-1}{x}}{\dfrac{\sqrt[3]{1+x}-1}{x}-\dfrac{\sqrt{1+x}-1\strut}{x}}$$ and use the definition of the derivative.