Limit evaluation, de l'Hopital seems not working

Question

I have to evaluate \begin{equation} L=\lim_{x\to 0} \frac{1-e^{\frac{x^2}{2}}\cos x}{2\sin^2x -x \arctan2x} \end{equation}

I tried with de l'Hopital's rule but it seems not working, am I missing something?

EDIT: I'll explain better what I tried & where I got stuck.
Straightforward substitution leads to a $\frac{0}{0}$ form. So I tried de l'Hopital:

Let $f(x)=1-e^{\frac{x^2}{2}}\cos x$ (the numerator) and $g(x)=2\sin^2x -x \arctan2x$ (the denominator).

We have:

\begin{equation} f'(x) = e^{\frac{x^2}{2}}(\sin x -x \cos x) \end{equation}

\begin{equation} g'(x) = -\frac{2x}{1+4x^2}-\arctan(2x) +4\sin x \cos x \end{equation}

This still leads to a $\frac{0}{0}$ form.

Differentiating again, we have

\begin{equation} f''(x) = xe^{\frac{x^2}{2}}(2\sin x -x \cos x) \end{equation}

\begin{equation} g''(x)= 4\left(-\sin^2 x +\cos^2 x -\frac{1}{(4x^2+1)^2} \right) \end{equation}

Still, evaluating the limit again leads to a $\frac{0}{0}$ form. Am I losing something?

Answer 1

Actually, if you want to use de l'Hopital's rule, you have to tediously calculate derivatives up to the fourth order. If we denote with $f(x)=1-e^\frac{x^2}{2}\cos(x)$ and $g(x)=2\sin^2(x)-x\arctan(2x)$, we have that: $$f^{(4)}(x)=e^\frac{x^2}{2}\left[(4x^3+8x)\sin(x)+(2-x^4)\cos(x)\right]$$ and $$g^{(4)}(x)=16\sin^2(x)-16\cos^2(x)+\frac{64}{(4x^2+1)^2}-\frac{1792x^2}{(4x^2+1)^3}+\frac{6144x^4}{(4x^2+1)^4}$$ The hypothesis of de l'Hopital's rule allow us to conclude that: $$\lim_{x\to 0}\frac{f(x)}{g(x)}=\lim_{x\to 0}\frac{f^{(4)}(x)}{g^{(4)}(x)}=\frac{2}{-16+64}=\frac{1}{24}$$

Edit: As pointed out in the comments, we first have to show that the limit $\lim_{x\to 0}\frac{f^{(3)}(x)}{g^{(3)}(x)}$ is in the $0/0$ form. Writing that we have to tediously calculate derivatives up to the fourth order, in order to make my answer more concise, I implicitly assumed, wrongly, that this fact could be easily shown. Obviously I shoud have specified it. Anyway... We have that: $$f^{(3)}(x)=e^\frac{x^2}{2}\left[(3x^2+2)\sin(x)-x^3\cos(x)\right]$$ and $$g^{(3)}(x)=-16\cos(x)\sin(x)+\frac{64x}{(4x^2+1)^2}-\frac{256x^3}{(4x^2+1)^3}$$ So $$\lim_{x\to 0}\frac{f^{(3)}(x)}{g^{(3)}(x)}$$ is in the form $0/0$. The fact that all the limits $\lim_{x\to 0}\frac{f^{(k)}(x)}{g^{(k)}(x)}$, for $k\leq 3$, are in the indeterminate form $0/0$ and that we can, finally, apply de l'Hopital's rule when $k=4$, is a consequence, as shown by other users, that the McLaurin series of $f(x)$ and $g(x)$ both have the first non-zero term of order $4$.

Answer 2

Using Taylor Series Expansion can be an efficient approach in this situation.

Expansion of the Numerator $ f(x) $:

• The Taylor series expansion of the numerator $ f(x) = 1 - e^{\frac{x^2}{2}} \cos x $ up to the sixth order around $ x = 0 $ is $ \frac{x^4}{12} $.

Expansion of the Denominator ( g(x) ):

• For $ \sin^2 x $, the expansion up to the sixth order around $ x = 0 $ is $ x^2 - \frac{x^4}{3} $.
• For $ \arctan(2x) $, the expansion up to the sixth order around $ x = 0 $ is $ 2x - \frac{4x^3}{3} + \frac{8x^5}{5} $.
• Combining these, the Taylor series expansion of the denominator $ g(x) = 2\sin^2 x - x \arctan(2x) $ up to the sixth order around $ x = 0 $ is $ 2(x^2 - \frac{x^4}{3}) - x(2x - \frac{4x^3}{3} + \frac{8x^5}{5}) $.

Simplified Limit Expression:

• With these expansions, the limit simplifies to $$ \lim_{x\to 0} \frac{\frac{x^4}{12}}{2(x^2 - \frac{x^4}{3}) - x(2x - \frac{4x^3}{3} + \frac{8x^5}{5})} = \frac{1}{24} $$

Answer 3

As an alternative, to avoid high order derivatives using standard limits and/or l'Hopital for simpler expressions, we can use that

$$\frac{1-e^{\frac{x^2}{2}}\cos x}{2\sin^2x -x \arctan2x}=\frac{1-e^{\frac{x^2}{2}}\cos x}{x^4}\frac{x^4}{2\sin^2x -x \arctan2x} \to -\frac1{12}\cdot \frac 12=-\frac1{24}$$

indeed for the first factor

$$\frac{1-e^{\frac{x^2}{2}}\cos x}{x^4}=\frac12\frac{e^{\frac{x^2}{2}}-1}{\frac {x^2}2}\frac{1-\cos x}{x^2}+\frac{2-e^{\frac{x^2}{2}}-\cos x}{x^4} \to\frac14-\frac16= -\frac1{12}$$

indeed using that $\frac{e^x-x-1}{x^2}\to \frac12$ and $\frac{\sin x -x}{x^3}\to -\frac16$

$$\frac{2-e^{\frac{x^2}{2}}-\cos x}{x^4}=-\frac14\frac{e^{\frac{x^2}{2}}-\frac{x^2}{2}-1}{\frac{x^4}4}-\frac{\cos x+\frac{x^2}{2}-1}{x^4}\to -\frac18-\frac1{24}=-\frac1{6}$$

since for the second one

$$\lim_{x\to 0}\frac{\cos x+\frac{x^2}{2}-1}{x^4}\stackrel{H.R.}=\lim_{x\to 0}\frac14\frac{x-\sin x}{x^3}=\frac1{24}$$

and for the second factor using that $\frac{x-\arctan x}{x^3}\to \frac13$ $$\frac{2\sin^2x -x \arctan2x}{x^4}=\frac{2\sin^2x -2x^2+2x^2-x \arctan2x}{x^4}=$$

$$=2\frac{\sin x -x}{x^3}\frac{\sin x +x}{x}+8\frac{2x- \arctan2x}{(2x)^3}\to-\frac23+\frac{8}3=2$$

Answer 4

It is better to first deal with denominator and we can write it as $$2(\sin^2x-x^2)+x(2x-\arctan 2x)$$ which can be further written as $$2(\sin x-x) (\sin x+x) +x(2x-\arctan 2x)\tag{1}$$ The above split helps us to evaluate the overall limit very easily. The key is to bring the following limit $$\lim_{x\to 0}\frac{\sin x-x} {x^3}=-\frac{1}{6}\tag{2}$$ into picture. The above limit is a standard exercise in many calculus texts and is easily handled by a single application of l'Hospital's Rule.

From $(2)$ we can observe that denominator $(1)$ can be divided by $x^4$ to get an expression with a finite limit. And this expression is $$2\cdot\frac{\sin x-x} {x^3}\cdot\left (\frac{\sin x}{x} +1\right) +\frac{2x-\arctan 2x}{x^3}\tag{3}$$ The first term above tends to $2\cdot (-1/6)\cdot 2=-2/3$. For the second term let's put $2x=\tan t$ to get the expression $$\frac{\tan t-t} {t^3}\cdot\frac{t^3}{\tan^3 t}\cdot 8$$ The first factor is handled by l'Hospital's Rule and the limit of the above expression is $(1/3)\cdot 1\cdot 8=8/3$. It now follows that the expression $(3)$ tends to $2=B\, \text{(say)} $.

Next we find the limit $A$ of the expression obtained by dividing numerator by $x^4$ namely $$\frac{1-e^{x^2/2}\cos x} {x^4}\tag{4}$$ To deal with this expression we make use of two well known limits $$\lim_{t\to 0}\frac{e^t-1}{t}=1,\lim_{x\to 0}\frac{1-\cos x} {x^2}=\frac{1}{2}\tag{5}$$ Replacing $t$ with $x^2/2$ in the first limit above we get $$\lim_{x\to 0}\frac{e^{x^2/2}-1}{x^2}=\frac{1}{2}$$ Multiplying this with second limit in $(5)$ we get $$\lim_{x\to 0}\frac {(e^{x^2/2}-1)(1-\cos x)} {x^4}=\frac{1}{4}$$ or $$\lim_{x\to 0}\frac {e^{x^2/2}-e^{x^2/2}\cos x-1+\cos x} {x^4}=\frac{1}{4}\tag{6}$$ We can now write $(4)$ as $$\frac{e^{x^2/2}-e^{x^2/2}\cos x-1+\cos x} {x^4}+\frac{2-\cos x-e^{x^2/2}}{x^4}$$ The first fraction tends to $1/4$ (via $(6)$) and to deal with second fraction we apply L'Hospital's Rule to get the expression $$\frac{\sin x-xe^{x^2/2}}{4x^3}$$ This can be further rewritten as $$\frac{1}{4}\cdot\left (\frac{\sin x - x} {x^3}-\frac{e^{x^2/2}-1}{x^2}\right)$$ and this tends to $$\frac{1}{4}\left(-\frac {1}{6}-\frac{1}{2}\right)=-\frac{1}{6}$$ It follows that limit $A$ of expression $(4)$ is $1/4-1/6=1/12$.

The desired limit of expression in question is thus $A/B=1/24$. In the overall procedure we should note that we have applied l'Hospital's Rule three times and each application is easy to handle with no complicated expressions being generated.

In general when applying l'Hospital's Rule it is better to first do a bit of algebraic manipulation. Also it should be noted that the cost of algebraic manipulation (including figuring them out) and l'Hospital's Rule exceeds the cost of Taylor series approach (for current problem) by a wide margin and hence it is preferable to opt for Taylor expansions here.