$\lim_{x \rightarrow 0} \frac{(x^2 - x^3) ((x + 1)^\sqrt{2} - 1)}{\log(1 + 4 x) (1 - \cos(2 x))}$

Question

I'm trying to find the following limit:

$$\lim_{x \rightarrow 0} \frac{(x^2 - x^3) ((x + 1)^\sqrt{2} - 1)}{\log(1 + 4 x) (1 - \cos(2 x))}$$

Knowing that:

$$ \lim_{x \rightarrow 0} \frac{\log(1 + x)}{x} = 1$$

and

$$ \lim_{x \rightarrow 0} \frac{1 - \cos(x)}{\frac{x^2}{2}} = 1$$

We have:

$$\lim_{x \rightarrow 0} \frac{(x^2 - x^3) ((x + 1)^\sqrt{2} - 1)}{\log(1 + 4 x) (1 - \cos(2 x))} = \lim_{x \rightarrow 0} \frac{1}{(4 x) \frac{(2 x)^2}{2}} \frac{(x^2 - x^3) ((x + 1)^\sqrt{2} - 1)}{\frac{\log(1 + 4 x)}{4 x} \frac{(1 - \cos(2 x))}{\frac{(2x)^2}{2}}} = \lim_{x \rightarrow 0} \frac{(x^2 - x^3) ((x + 1)^\sqrt{2} - 1)}{(4 x) \frac{(2 x)^2}{2}} = \lim_{x \rightarrow 0} \frac{(1 - x)((x + 1)^\sqrt{2} - 1)}{8 x} $$

I feel I need to get rid of $((x + 1)^\sqrt{2} - 1)$ in numerator (so that I don't have a $\frac{0}{0}$ indeterminate form), but I don't know how.

Edit:

Knowing that:

$$ \lim_{x \rightarrow 0} \frac{e^x - 1}{x} = 1 $$

We have that:

$$\lim_{x \rightarrow 0} (x + 1)^\sqrt{2} - 1 = \lim_{x \rightarrow 0} \frac{e^{\sqrt{2} \log(x + 1)} - 1}{\sqrt{2} \log(x + 1)} = \lim_{x \rightarrow 0} \frac{e^{\sqrt{2} \log(x + 1)} - 1}{\sqrt{2} \log(x + 1)} \frac{\log(x + 1)}{x} \sqrt{2} x$$

I think that should solve it.

Answer 1

Yes you idea by standard limits is fine, after factorization

$$\frac{(x^2 - x^3) ((x + 1)^\sqrt{2} - 1)}{\log(1 + 4 x) (1 - \cos(2 x))}=\frac{(x+1)^\sqrt2-1}x\cdot\frac{x^2}{1-\cos2x}\cdot\frac{4x}{\log(1+4x)}\cdot\frac{1-x}{4}$$

for the first factor we have

$$\frac{(x+1)^\sqrt2-1}x=\sqrt 2\; \frac{e^{\sqrt 2 \log(x+1)}-1}{\sqrt 2 \log(x+1)}\frac{\log(x+1)}{x} \to \sqrt 2\cdot 1 \cdot 1=\sqrt 2$$

Answer 2

This is one using series expansions for elementary functions

$$ \lim_{x \rightarrow 0} \frac{(x^2 - x^3) ((x + 1)^\sqrt{2} - 1)}{\log(1 + 4 x) (1 - \cos(2 x))}=\lim_{x \rightarrow 0} \frac{x^2 ((x + 1)^\sqrt{2} - 1)}{\log(1 + 4 x) (1 - \cos(2 x))} $$

$$ \\ \ \\ \ \\ (1+x)^n=1+nx+{n(n-1)\over 2!}x^2+\cdots\\ \log(1+x)=x-{x^2\over 2}+{x^3\over 3}+\cdots\\ \cos x=1-{x^2\over2!}+{x^4\over4!}+\cdots $$

So the limit becomes

$$ =\lim_{x \rightarrow 0} \frac{x^2 (1+\sqrt{2}x+{\sqrt{2}(\sqrt{2}-1)x^2\over2!}\cdots - 1)}{(4x-{16x^2\over2}+\cdots) (1 - (1-{4x^2\over2!}+{16x^4\over4!}+\cdots))}\\ =\lim_{x \rightarrow 0} \frac{x^3 (\sqrt{2}+{\sqrt{2}(\sqrt{2}-1)x\over2!}\cdots )}{x^3(4-{16x\over2}+\cdots) ({4\over2!}+{16x^2\over4!}+\cdots))}={\sqrt{2}\over8} $$

Answer 3

$$\frac{(x^2 - x^3) ((x + 1)^\sqrt{2} - 1)}{\log(1 + 4 x) (1 - \cos(2 x))}=\frac{(x+1)^\sqrt2-1}x\cdot\frac{x^2}{1-\cos2x}\cdot\frac{x-x^2}{\log(1+4x)}\;(**)$$

and just as Paramanand told you twice, using

$$\lim_{x\to0}\frac{(x+1)^\sqrt2-1}x=\left[\left((x+1)^\sqrt2\right)'\right]_{x=0}=\sqrt2\,(0+1)^{\sqrt2-1}=\sqrt2$$

we get at the end

$$(**)\xrightarrow[x\to0]{}\sqrt2\cdot\frac12\cdot\frac14=\frac{\sqrt2}8=\frac1{4\sqrt2}$$