3
$\begingroup$

Evaulate $$ L = \lim_{x \to 0} \frac{1-\cos(\sin x)+\ln(\cos x)}{x^4}. $$

I can solve it using Maclaurin series, but I'm trying to figure out a way to the solution without using it. L'Hopital would probably work but needs to be applied 4 times, and the given function in numerator is too complicated for things to work out nicely on paper.

I think we can somehow use $\displaystyle\lim_{x\to0}\frac{1-\cos(x)}{x^2}=\frac12$ and $\displaystyle\lim_{x\to0}\frac{\ln(1+x)}{x}=1$.

With these, I can lower the degree of denominator to $2$, like this:

$$ L = \lim_{x \to 0}\frac{\frac{1 - \cos(\sin x)}{\sin^2 x}\frac{\sin^2 x}{x^2}+\frac{\ln(1+(\cos x-1))}{\cos x - 1}\frac{\cos x - 1}{x^2}}{x^2}. $$

This seems closer because left and right terms in the numerator are made of known limits, but the problem is that the whole expression is still $(1/2 - 1/2)/0 = 0/0$ so I can not break down $L$ into two limits and work things out.

I'm looking only for solutions involving manipulating limits in this way. No Maclaurin and no L'Hopital if it's gonna get too messy on paper.

$\endgroup$

5 Answers 5

1
$\begingroup$

Instead of using the usual limits
$$ \lim_{y\rightarrow 0}\frac{1-\cos y}{y^{2}}=\frac{1}{2},\ \ \ \ \ \ \ \ \lim_{y\rightarrow 0}\frac{\log (1+y)}{y}=1. $$ which lead to $$ L=\lim_{x\rightarrow 0}\frac{\frac{1-\cos (\sin x)}{\sin ^{2}x}\frac{\sin ^{2}x}{x^{2}}+\frac{\ln (1+(\cos x-1))}{\cos x-1}\frac{\cos x-1}{x^{2}}}{% x^{2}}. $$ I will push one more order and use the (maybe less usual but easy to obtain) following limits $$ \lim_{y\rightarrow 0}\frac{1-\frac{y^{2}}{2}-\cos y}{y^{4}}=-\frac{1}{24},\ \ \ \ \ and \ \ \ \ \ \ \ \lim_{y\rightarrow 0}\frac{\log (1-y)+y}{y^{2}}=-\frac{1}{2}. $$

Each one of them maybe obtained by making use of L'Hospital's rule. \begin{eqnarray*} L &=&\lim_{x\rightarrow 0}\frac{1-\cos \left( \sin x\right) +\ln \left( \cos x\right) }{x^{4}} \\ &=&\lim_{x\rightarrow 0}\frac{1-\frac{\sin ^{2}\left( x\right) }{2}-\cos \left( \sin x\right) +\frac{\sin ^{2}\left( x\right) }{2}+\ln \left( \cos x\right) }{x^{4}} \\ &=&\lim_{x\rightarrow 0}\left( \frac{1-\frac{\sin ^{2}\left( x\right) }{2}% -\cos \left( \sin x\right) }{\sin ^{4}x}\right) \left( \frac{\sin x}{x}% \right) ^{4}+\frac{\frac{\sin ^{2}\left( x\right) }{2}+\ln \left( \sqrt{% 1-\sin ^{2}x}\right) }{x^{4}} \\ &=&\lim_{x\rightarrow 0}\left( \frac{1-\frac{\sin ^{2}\left( x\right) }{2}% -\cos \left( \sin x\right) }{\sin ^{4}x}\right) \left( \frac{\sin x}{x}% \right) ^{4}+\frac{1}{2}\frac{\sin ^{2}x+\ln \left( 1-\sin ^{2}x\right) }{% \left( \sin ^{2}x\right) ^{2}}\left( \frac{\sin x}{x}\right) ^{4} \\ &=&\lim_{y\rightarrow 0}\left( \frac{1-\frac{y}{2}-\cos \left( y\right) }{% y^{4}}\right) \cdot \left( 1\right) ^{4}+\frac{1}{2}\lim_{y\rightarrow 0}% \frac{y+\ln \left( 1-y\right) }{y^{2}}\cdot \left( 1\right) ^{4} \\ &=&-\frac{1}{24}+\frac{1}{2}\cdot \frac{-1}{2}=-\frac{7}{24}. \end{eqnarray*}

$\endgroup$
0
$\begingroup$

Let $x=2\theta$, then:

$1-\cos(\sin x) \approx_0 \dfrac{\sin^2 x}{2}=\dfrac{\sin^2(2\theta)}{2}$

$\log(\cos x) \approx_0 \cos x - 1=-2\sin^2\left(\frac{x}{2}\right)=-2\sin^2\theta$.

Thus $L = \displaystyle \lim_{\theta \to 0} \dfrac{\dfrac{\sin^2(2\theta)}{2}-2\sin^2\theta}{16\theta^4}=-\dfrac{1}{8}\displaystyle \lim_{\theta \to 0} \left(\dfrac{\sin\theta}{\theta}\right)^4=-\dfrac{1}{8}$

$\endgroup$
1
  • 2
    $\begingroup$ You can't ignore the fourth-order terms if you divide by $x^4$. $\endgroup$ Commented May 15, 2015 at 22:39
0
$\begingroup$

Using L'Hopital partially three times gives you$$\lim_{x\to 0}\frac{1-\cos(\sin x)+\ln(\cos x)}{x^4}$$$$=\lim_{x\to 0}\frac{1}{4\cos x}\lim_{x\to 0}\frac{\sin(\sin x)\cos^2x-\sin x}{x^3}$$$$=\frac 14\left(\lim_{x\to 0}\frac{-2\cos x}{3}\cdot\left(\frac{\sin x}{x}\right)^2\cdot \frac{\sin (\sin x)}{\sin x}+\frac 13\lim_{x\to 0}\frac{\cos(\sin x)\cos^3x-\cos x}{x^2}\right)$$$$\small{=\frac 14\left(-\frac 23\right)+\frac{1}{4\cdot 3}\lim_{x\to 0}\left(-\frac{\sin(\sin x)}{\sin x}\cdot \frac{\sin x}{x}\cdot\frac{\cos^4x}{2}-\frac{3\cos(\sin x)\cos^2x}{2}\cdot\frac{\sin x}{x}+\frac 12\cdot\frac{\sin x}{x}\right)}$$$$=\color{red}{-\frac{7}{24}}$$

$\endgroup$
0
$\begingroup$

We can let $x = \arcsin y$ and take the limit as $y\to 0.$ Note that $\cos (\arcsin y) = (1-y^2)^{1/2}.$ So we are looking at

$$(1)\,\,\,\,\lim_{y\to 0} \frac{1 - \cos y + \ln (1-y^2)^{1/2}}{(\arcsin y)^4}.$$

Now $(\arcsin y)/y \to 1,$ so we can replace the denominator in (1) by $y^4.$

Time for some Taylor: $\cos y = 1 - y^2/2 + y^4/4! +O(y^6).$ Also, for small $h> 0$ we have $\ln (1-h)= -(h+h^2/2 + O(h^3)).$ Thus $$\ln (1-y^2)^{1/2} = (1/2)\ln (1-y^2) = -(1/2)(y^2 + y^4/2 +O(y^6)).$$ Put all of this into (1), and we get some nice cancellation to obtain the limit of $-7/24.$

$\endgroup$
0
$\begingroup$

We have $$1-\cos(\sin(x)) = 1 - \left(\sum_{k=0}^{\infty} (-1)^k\dfrac{\sin^{2k}(x)}{(2k)!}\right)$$ $$\ln(\cos(x)) = \dfrac12 \ln\left(\cos^2(x)\right) = \dfrac{\ln(1-\sin^2(x))}2 = -\dfrac12 \sum_{k=1}^{\infty} \dfrac{\sin^{2k}(x)}k$$ Hence, we have the function to be \begin{align} f(x) & = \dfrac{-\displaystyle\sum_{k=1}^{\infty}(-1)^k \dfrac{\sin^{2k}(x)}{(2k)!} - \displaystyle\sum_{k=1}^{\infty}\dfrac{\sin^{2k}(x)}{(2k)}}{x^4} = \dfrac{\displaystyle\sum_{k=2}^{\infty}\sin^{2k}(x) \left(\dfrac{(-1)^{k-1}-(2k-1)!}{(2k)!}\right)}{x^4}\\ & = \dfrac{\displaystyle\sum_{k=2}^{\infty} c_k \sin^{2k}(x)}{x^4} \end{align} where $c_k = \dfrac{(-1)^{k-1}-(2k-1)!}{(2k)!}$. Hence, $$\lim_{x \to 0}f(x) = c_2 = \dfrac{(-1)^{2-1}-(2\cdot2-1)!}{(2\cdot2)!} = \dfrac{-1-3!}{4!} = -\dfrac7{24}$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .