Solving this limit with Taylor series

Question

I have this limit which I "must" solve with Taylor series but I don't really know where I am wrong and how the result comes out...

$$\lim_{x\to 0^+} \frac{\ln(1 + \sqrt{1-\cos(x)})}{1-e^{\sin(3x)}}$$

The result is $-\frac{1}{3\sqrt{2}}$.

Now here is what I have done, even if it's a disaster.

Starting with the denominator, I expand the exponential, and then the argument up to second orders:

$$ \begin{align} e^{\sin(3x)} & = 1 + \sin(3x) + \frac{1}{2}\sin^2(3x) + o(\sin^2(3x)) \\\\ & = 1 + \left[3x + o(x^2)\right] + \frac{1}{2}\left[x + o(x)\right]^2 + o (x^2) \\\\ & = 1 + 3x + \frac{1}{2}x^2 + o(x^2) \end{align} $$

Now the logarithm: I use Taylor series for the logarithm

$$ \begin{align} \ln(1 + \sqrt{1-\cos(x)}) = \sqrt{1-\cos(x)} - \frac{1}{2}(1-\cos(x)) + o(1-\cos(x)) \end{align} $$

Then I expand by using $(1 + z)^{\alpha}$ where $\alpha = 1/2$ and lastly the cosine:

$$ \begin{align} & \Big(1 - \frac{1}{2}\cos(x) + \frac{3}{8}\cos^2(x) + o(\cos^2(x))\Big) - \frac{1}{2} + \frac{1}{2}\cos(x) + o(x^2) \\\\ & = \frac{1}{2} + \frac{3}{8}\cos^2(x) + o(x^2) \\\\ & = \frac{1}{2} + \frac{3}{8}\left(1 - \frac{x^2}{2} + o(x^2)\right)^2 + o(x^2) \\\\ & = \frac{1}{2} + \frac{3}{8} \left(1 - x^2 + o(x^2)\right) \\\\ & = \frac{7}{8} + \frac{3x^2}{8} + o(x^2) \end{align} $$

Now when substituting:

$$\lim_{x\to 0^+} \frac{\ln(1 + \sqrt{1-\cos(x)})}{1-e^{\sin(3x)}} = \lim_{x\to 0^+} \frac{\frac{7}{8} + \frac{3x^2}{8} + o(x^2)}{-3x - \frac{1}{2}x^2 + o(x^2)}$$

Using the $o(x)$ algebra, hence collecting $x^2$ so $\frac{o(x^2)}{x^2} \to 0$:

$$\lim_{x\to 0^+} \frac{x^2\left(\frac{7}{8x^2} + 3 + \frac{o(x^2)}{x^2}\right)}{x^2 \left(-\frac{3}{x} - \frac{1}{2} + \frac{o(x^2)}{x^2}\right)} = \lim_{x\to 0^+} \frac{\frac{7}{8x^2} + 3}{-\frac{3}{x}-\frac{1}{2}} = \lim_{x\to 0^+} \frac{\frac{7 + 24 x^2}{8x^2}}{\frac{-6-x}{2x}}$$

This surely doesn't go to $-\frac{1}{3\sqrt{2}}$.

I tried this really $7-8$ times and I keep getting wrong resuslts.

Any help? Thank you!

Answer 1

When you are doing Taylor expansions, it is very important to stick with one $o(f(x))$ and stick to it, meaning eliminating everything negligible compared to $f(x)$.

Here you have some mixed $o(\sin(x)^2)$ and $o(1-\cos(x))$, it would be better to expand them in $o(x^n)$ before proceeding to $\exp,\ln$ and $\sqrt{}$ expansions.

Note that I do not use $|x|$ below since $x\to 0^+$, so I use $x$ instead.

• numerator

$\begin{align}\ln(1+\sqrt{1-\cos(x)}) &=\ln(1+\sqrt{\frac 12x^2+o(x^2)})\\ &=\ln(1+\frac x{\sqrt{2}}\sqrt{1+o(1)})\\ &=\ln(1+\frac x{\sqrt{2}}(1+o(1))\\ &=\ln(1+\frac x{\sqrt{2}}+o(x))\\ &=\frac x{\sqrt{2}}+o(x) \end{align}$

• denominator

$\begin{align}1-\exp(\sin(3x))&=1-\exp(3x+o(x))\\&=1-(1+3x+o(x))\\&=-3x+o(x)\end{align}$

And you get a limit in $-\frac 1{3\sqrt{2}}$

Answer 2

Note that $$\lim_{x\to 0+}\frac{\ln(1+\sqrt{1-\cos(x)})}{\sqrt{1-\cos(x)}}=\lim_{y\to 0+}\frac{\ln(1+y)}{y}=1.\tag 1$$ The second equality may be obtained using series for $\ln(1+y)$.

$\sqrt{1-\cos(x)}=\sqrt 2 \sin(x/2)$ for small $x>0$.

$\begin{align}1-e^{\sin 3x}&=-\sin(3x)-1/2 \sin^2(3x)- o(\sin^2 (3x))\\&=-(3\sin x- 4\sin^3x)(1+1/2\sin(3x)+o(\sin(3x)))\end{align}$

So using $\sin (x)= 2\sin(x/2)\cos(x/2)$ gives$$\lim_{x\to 0+}\frac{\sqrt{1-\cos(x)}}{1-e^{\sin 3x}}= \lim_{x\to 0+}\frac {\sqrt 2} {-2\cos(x/2)(3-\sin^2x)(1+1/2\sin(3x)+o(\sin(3x)))}=-\frac 1{3\sqrt 2}.\tag 2$$

$\lim_{x\to 0+}\frac{\ln(1+\sqrt{1-\cos(x)})}{1-e^{\sin 3x}}=\left(\lim_{x\to 0+}\frac{\ln(1+\sqrt{1-\cos(x)})}{\sqrt{1-\cos(x)}}\right)\left(\lim_{x\to 0+}\frac{\sqrt{1-\cos(x)}}{1-e^{\sin 3x}}\right)=-\frac 1{3\sqrt 2}$ by $(1)$ and $(2)$.

Answer 3

$$\frac{\log(1 + \sqrt{1-\cos(x)})}{1-e^{\sin(3x)}}$$

Using pieces $$\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{24}+O\left(x^6\right)$$ $$1-\cos(x)=\frac{x^2}{2}-\frac{x^4}{24}+O\left(x^6\right)$$

Assuming $x>0$ $$\sqrt{1-\cos(x)})=\frac{x}{\sqrt{2}}-\frac{x^3}{24 \sqrt{2}}+O\left(x^4\right)$$ $$\log(1 + \sqrt{1-\cos(x)})=\frac{x}{\sqrt{2}}-\frac{x^2}{4}+\frac{x^3}{8\sqrt{2}}+O\left(x^4\right)$$ $$\sin(3x)=3 x-\frac{9 x^3}{2}+O\left(x^4\right)$$ $$e^{\sin(3x)}=1+3 x+\frac{9 x^2}{2}+O\left(x^4\right)$$

So, we have $$\frac {\frac{x}{\sqrt{2}}-\frac{x^2}{4}+\frac{x^3}{8\sqrt{2}}+O\left(x^4\right) } {-3 x-\frac{9 x^2}{2}+O\left(x^4\right) }$$

Long division $$-\frac{1}{3 \sqrt{2}}+\frac{1+3 \sqrt{2}}{12} x-\frac{6+19 \sqrt{2}}{48} x^2+O\left(x^3\right)$$

Do the same using $x<0$ for a different result.

When you use Taylor, I suggest to work from inside to outside.