Calculating function limits using infinitesimals

Question

So I have two examples:

1. $ \displaystyle\lim_{x \to 0} \frac{\cos(xe^x) - \cos(xe^{-x})}{x^3} $
2. $\displaystyle\lim_{x \to 0} \frac{xe^{-\frac{x}{2}} - \sin(\ln(1+x))}{x^3}$

Using the same rules:

$e^x \sim 1+x$

$\sin(x) \sim x$

$\ln(1+x) \sim x$

$\cos(x) \sim 1 - \frac{x^2}{2}$

I get $-2$ for the first limit, which is the right answer, but I get $+\infty$ for the second one, but that answer is wrong. Can someone explain me how this works or tell me where to look up more about this way of solving limits. It seems that I'm missing something important here.

The way I solved them is just by replacing $\sin(x)$ with $x$ and so on.

Answer 1

The infinitesimals you have used are insufficient for the numerator of the second problem. Both \begin{align*} \left. \left( \frac{\mathrm{d}}{\mathrm{d}x} \left( x \mathrm{e}^{-x/2} - \sin(\ln(x+1)) \right) \right) \right|_{x \rightarrow 0} &= 0 \text{ and } \\ \left. \left( \frac{\mathrm{d}^2}{\mathrm{d}x^2} \left( x \mathrm{e}^{-x/2} - \sin(\ln(x+1)) \right) \right) \right|_{x \rightarrow 0} &= 0 \text{.} \end{align*} So you need to raise the order of approximation. \begin{align*} \mathrm{e}^x &\sim 1+x+\frac{x^2}{2} &&\text{,} \\ \sin(x) &\sim x - \frac{x^3}{6} &&\text{, and } \\ \ln(1+x) &\sim x - \frac{x^2}{2} + \frac{x^3}{3} \text{,} \end{align*} where we do not extend the series for sine because it is of third order. Notice we only extend enough to ensure that we get all terms in $x^3$ and of lower degree. To see that:

• $\mathrm{e}^x$ has degree $2$ and so does $\mathrm{e}^{-x/2}$. Multiplying by $x$ gives us degree $3$.
• The "$x$" in the replacement for "$\sin x$" necessitates keeping terms to degree $3$ in $\ln(1+x)$. The "$x$" in the replacement for "$\ln(1+x)$" necessitates keeping the $x^3$-term in sine.

Answer 2

Observe that : $ \left(\forall x\in\mathbb{R}^{*}\right),\ \large \frac{x\,\mathrm{e}^{-\frac{x}{2}}-\sin{\left(\ln{\left(1+x\right)}\right)}}{x^{3}}=\frac{\mathrm{e}^{-\frac{x}{2}}-1+\frac{x}{2}}{x^{2}}-\frac{\ln{\left(1+x\right)}-x+\frac{x^{2}}{2}}{x^{3}}+\frac{\ln{\left(1+x\right)}-\sin{\left(\ln{\left(1+x\right)}\right)}}{\ln^{3}{\left(1+x\right)}}\left(\frac{\ln{\left(1+x\right)}}{x}\right)^{3} $

Now all that remains to do is to determine the value of each of the following well-known limits :

$$ \lim_{x\to 0}{\frac{x-\sin{x}}{x^{3}}} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lim_{x\to 0}{\frac{\mathrm{e}^{x}-1-x}{x^{2}}} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lim_{x\to 0}{\frac{\ln{\left(1+x\right)}-x+\frac{x^{2}}{2}}{x^{3}}}$$

If you'd like further help, feel free to ask.

Answer 3

Limits are evaluated using theorems meant to evaluate them. These include laws of algebra of limits, squeeze theorem, L'Hospital's Rule, Taylor expansions and many more for specialized problems. Also one can use a set of well known standard limit formulas along with these theorems.

What you have done is not based on any of the above theorems and thus it is not guaranteed to work. Especially you should not replace one expression by another (like $\sin x$ by $x$) unless they are equal.

For the first problem you can simply use the formula $$\cos a - \cos b=2\sin\frac{a+b}{2}\sin\frac{b-a}{2}$$ and thus the expression under limit becomes $$(-2) \cdot\frac{\sin (x\cosh x) \sin(x\sinh x) } {x^3}$$ which can be further rewritten as $$(-2) \cdot\frac{\sin(x\cosh x)} {x\cosh x} \cdot \cosh x\cdot\frac{\sin(x\sinh x)} {x\sinh x} \cdot\frac{\sinh x} {x} $$ and the desired limit is thus $(-2)$ as all other factors tend to $1$.

The second problem is slightly complicated and necessitates the use of tools like L'Hospital's Rule or Taylor's theorem. Here is one approach via L'Hospital's Rule. Split the numerator by adding and subtracting $\log(1+x)$ to get the expression under limit as $$\frac{xe^{-x/2}-\log(1+x)}{x^3}+\frac{\log(1+x)-\sin\log(1+x)}{(\log(1+x))^3}\cdot\left(\frac{\log(1+x)}{x}\right)^3$$ Using the substitution $t=\log(1+x)$ the second term becomes $$\lim_{t\to 0}\frac{t-\sin t} {t^3}=\frac{1}{6}$$ via L'Hospital's Rule. So the desired limit is $$\frac{1}{6}+\lim_{x\to 0}\frac{xe^{-x/2}-\log(1+x)}{x^3}$$ This one can be handled easily via Taylor series using $$e^{-x/2}=1-\frac{x}{2}+\frac{x^2}{8}+o(x^2),\log(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}+o(x^3)$$ and one easily gets the desired limit as $$\frac{1}{6}-\frac{5}{24}=-\frac{1}{24}$$ While using Taylor expansions for limit evaluation one should observe that they are nothing more than a set of well known limit formulas for every well behaved function. And one should also see that one does not have to multiply, divide or compose two Taylor expansions unless absolutely necessary.