Solving the following limit without using neither L'Hopital's Rule nor series expansions

Question

I've been struggling for days on how to solve the following limit:

$\begin{gather*} \lim\limits_{x\to 0} \frac{x^3\sin(x)}{[\ln(1+x)-x]^2} \end{gather*}$

I shouldn't use neither the L'Hopital's rule nor series expansions.

I tried to simplify the limit by using the fact that:

$\begin{gather*} \lim\limits_{x\to 0} \frac{\sin(x)}{x} = 1 \end{gather*}$

...and I got:

$\begin{gather*} \lim\limits_{x\to 0} {\bigg(\frac{x^2}{\ln(1+x)-x}\bigg)^2} \end{gather*}$

To semplify the problem I simply tried to solve:

$\begin{gather*} \lim\limits_{x\to 0} {\frac{x^2}{\ln(1+x)-x}} \end{gather*}$

...remembering that the result of this limit should be squared to get the limit we're looking for.

Then I used the fact that $e^{\ln(x)}=x$ and so I obtained:

$\lim\limits_{x\to 0} {\frac{x^2}{e^{(\ln(\ln(1+x)-x))}}}$

At this point I did a substitution: $\ln(1+x) = t $ And so I got:

$\lim\limits_{x\to 0} {\frac{(e^t-1)^2}{e^{(\ln(t+1-e^t))}}}$

which is equal to:

$\lim\limits_{x\to 0} {\frac{(e^t-1)^2}{-e^t+t+1}}$ = $\lim\limits_{x\to 0} {-\frac{(e^t-1)^2}{e^t-t-1}}$

At this point I got stucked.

Any suggestion will be very welcomed, thanks in advance.

Simon

Answer 1

I am posting a new answer building on your work.

Let $0<x\le1$. The Binomial theorem gives, for all $n\ge3$, $$\left(1+\frac xn\right)^n=\sum_{k=0}^n\binom nk\frac{x^k}{n^k} =1+x+\frac{n(n-1)}2\frac{x^2}{n^2} +\sum_{k=3}^n\binom nk\frac{x^k}{n^k}.$$ Thus $$\left(1+\frac xn\right)^n\ge1+x+\frac{n(n-1)}2\frac{x^2}{n^2}$$ and (using that $x^k\le x^3$ for all $k\ge3$) \begin{align*} \left(1+\frac xn\right)^n&\le1+x+\frac{n(n-1)}2\frac{x^2}{n^2}+x^3\sum_{k=3}^n\binom nk\frac1{n^k}\\[.4em] &\le1+x+\frac{n(n-1)}2\frac{x^2}{n^2}+x^3\left(1+\frac1n\right)^n. \end{align*} Passing to the limit as $n\to\infty$ we obtain $$\mathrm e^x\ge1+x+\frac{x^2}2\qquad\text{and}\qquad\mathrm e^x\le1+x+\frac{x^2}2+\mathrm e\,x^3,$$ that is $$\frac{x^2}2\le\mathrm e^x-1-x\le\frac{x^2}2+\mathrm e\,x^3.$$ Dividing by $x^2>0$ we get $$\frac12\le\frac{\mathrm e^x-1-x}{x^2}\le\frac12+\mathrm e\,x.$$ Now taking the limit $x\to0^+$ shows, by the squeezing theorem, that $$\lim_{x\to0^+}\frac{\mathrm e^x-1-x}{x^2}=\frac12.$$

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Now write $$-\frac{(\mathrm e^t-1)^2}{\mathrm e^t-t-1}=-\left(\frac{\mathrm e^t-1}t\right)^2\cdot\frac{t^2}{\mathrm e^t-1-t}.$$

Answer 2

You can “replace” Taylor expansions by inequalities which may be obtained by simple integration or study of functions.

1. For all $t>0$, we have $(1-t)(1+t)=1-t^2\le1\le1+t^3=(1+t)(1-t+t^2)$, so $$1-t\le\frac1{1+t}\le1-t+t^2.$$ Integrating this inequality between $0$ and $x>0$ gives $$x-\frac{x^2}2\le\ln(1+x)\le x-\frac{x^2}2+\frac{x^3}3.$$ Thus, for all $0<x<\frac32$, $$0<\frac{x^2}2-\frac{x^3}3\le x-\ln(1+x)\le\frac{x^2}2,$$ and taking squares we obtain $$\fbox{$0<\frac{x^4}4-\frac{x^5}3+\frac{x^6}9\le\Bigl(x-\ln(1+x)\Bigr)^{\!2}\le\frac{x^4}4.$}\tag{1}$$
2. Similarly, if we integrate $|{\cos t}|\le1$ between $0$ and $x>0$ we have $$|{\sin(x)}|=\left\lvert\int_0^x\cos(t)\,\mathrm dt\right\rvert \le\int_0^x\lvert{\cos(t)}\rvert\,\mathrm dt\le\int_0^x1\,\mathrm dt=x.$$ Now integrating $|{\sin(t)}|\le t$ between $0$ and $x$ gives $$\lvert1-\cos(x)\rvert=\left\lvert\int_0^x\sin(t)\,\mathrm dt\right\rvert \le\int_0^x|{\sin(t)}|\,\mathrm dt\le\int_0^xt\,\mathrm dt=\frac{x^2}2.$$ Integrating one more time $\lvert1-\cos(t)\rvert\le\frac{t^2}2$ between $0$ and $x$ shows that $$|{\sin(x)}-x|=\left\lvert\int_0^x\bigl(1-\cos(t)\bigr)\,\mathrm dt\right\rvert\le\int_0^x|1-\cos(t)|\,\mathrm dt\le\int_0^x\frac{t^2}2\,\mathrm dt=\frac{x^3}6.$$ Thus for all $0<x<\frac1{\sqrt6}$, $$0<x-\frac{x^3}6\le\sin(x)\le x,$$ and, multiplying by $x^3>0$, $$\fbox{$0<x^4-\frac{x^6}6\le x^3\sin(x)\le x^4$.}\tag{2}$$
3. Taking $(1)$ over $(2)$ now yields $$4\le\frac{x^3\sin(x)}{\Bigl(\ln(1+x)-x\Bigr)^2}\le\frac{x^4}{\frac{x^4}4-\frac{x^5}3+\frac{x^6}9}.$$ for all $0<x<\frac1{\sqrt6}$. The right-hand side tends to $4$ as $x\to0^+$.