Solve the limit $\lim\limits_{x\to 0}\left(\frac {e^x}{x}-\frac {1}{\arctan{x}}\right)$ without using L'Hopital (with my attempts)

Question

Help me solve this limit, using simple limit work: $$\lim_{x\to 0}\left(\frac {e^x}{x}-\frac {1}{\arctan{x}}\right)$$

I tried extracting $\frac {e^x}{x}$ but that was dead end, then $\frac {e^x-1}{x}+\frac{1}{x}-\frac{1}{x} \frac{x}{\arctan{x}}$ but I got stuck there.

Use $$\dfrac{e^x-1}x$$ and https://math.stackexchange.com/questions/387333/are-all-limits-solvable-without-lh%c3%b4pital-rule-or-series-expansion — lab bhattacharjee, Jun 29 '19 at 20:32
Your approach is fine. The first term tends to $1$ and you show that remaining part tends to $0$. — Paramanand Singh, Jun 30 '19 at 00:53

score 5 · Answer 1 · answered Jun 29 '19 at 21:35

5

Everything is locally a power series, so, as $x \to 0$,

$\begin{array}\\ \left(\frac {e^x}{x}-\frac {1}{\arctan{x}}\right) &=\left(\frac {1+x+O(x^2)}{x}-\frac {1}{x-x^3/3+O(x^5)}\right)\\ &=\left(\frac {(1+x+O(x^2))(1-x^2/3+O(x^4))-1}{x(1-x^2/3+O(x^4))}\right)\\ &=\left(\frac {(1+x+O(x^2))-1}{x(1+O(x^2))}\right)\\ &=\left(\frac {x+O(x^2)}{x+O(x^3))}\right)\\ &=1+O(x)\\ \end{array} $

answered Jun 29 '19 at 21:35

marty cohen

107,799

This is unnecesserly complicated, and previous answer is better because it isn't using some general solving method but only wit. – dzaralica69 Jun 29 '19 at 21:39
2

So mine is half as good? – marty cohen Jun 30 '19 at 00:03
Twice as general so half as good?............+1 – DanielWainfleet Jun 30 '19 at 08:32

score 2 · Accepted Answer · answered Jun 29 '19 at 21:21

2

Rewrite everything as:

$ \Large{e^x\arctan x -x\over x \arctan x} $

$ \arctan x $ is asymptotically equal to $x$ as $x \to 0$ so we have:

$ \Large{e^x\arctan x -x\over x \arctan x}$ $\sim$ $\Large{{xe^x -x}\over{x^2}} $

Group by $x$ the numerator:

$\Large{{x(e^x -1)}\over{x^2}} $

$ e^x -1 $ is asymptotically equivalent to $x$ as $x \to 0$ so we have:

$\Large{{x(e^x -1)}\over{x^2}} $ $\sim$ $\Large{x^2\over x^2}$ $=$ $1$

answered Jun 29 '19 at 21:21

pitagorapp

94

Amazing! Since your good, can you give a look at this problem and solve it simply like this one:. https://math.stackexchange.com/q/3278072/685632 – dzaralica69 Jun 29 '19 at 21:26
1

That "asymptotically equivalent" sounds a little suspicious: what do you really mean by that? Unless clear rules are established about this, I guess you couldalso say that $;e^x;$ "is asymptotically equivalent" to $;1;$ when $;x\to0;$ , so in $;\cfrac{xe^x-x}{x^2}\sim\cfrac{x-x}{x^2}=0;$ ...so how does this work exactly? – DonAntonio Jun 29 '19 at 21:33
Asymptotically equivalent means that the limit as $x$ goes to $0$ of the ratio $e^x -1\over x$ is equal to 1, hence $e^x -1$ and $x$ are asymptotically equivalent near in a neighborhood of $0$ – pitagorapp Jun 29 '19 at 21:37
Asymptotic equivalences cant be used blindly. Thus replacing $\arctan x$ with $x$ in denominator is fine but not so in numerator. You need to evaluate the limit of $(e^x\arctan x - x) /x^2$. - 1 – Paramanand Singh Jun 30 '19 at 00:50

score 1 · Answer 3 · answered Jun 30 '19 at 08:09

Your own approach works fine. Just split the expression under limit as $$\frac{e^x-1}{x}+\frac{\arctan x-x} {x\arctan x} $$ The first fraction tends to $1$ and one show that the second fraction tends to $0$ so that the desired limit is $1$. Since $(\arctan x) /x\to 1$ the limit of second fraction above is same as that of $$\frac{\arctan x-x} {(\arctan x) ^2}$$ Putting $x=\tan t$ we see that $t\to 0$ and the above fraction reduces to $$\frac{t-\tan t} {t^2}$$ Clearly the limit of above expression is same as that of $$\frac{t\cos t-\sin t} {t^2}=\frac{\cos t - 1}{t}+\frac{t-\sin t} {t^2}$$ and you can easily prove that these two fractions tend to $0$.

To the proposer: You might notice that if $f(x)=e^x$ then $\lim_{x\to 0},(e^x-1)/x=f'(0)=1.$ — DanielWainfleet, Jun 30 '19 at 08:35

Solve the limit $\lim\limits_{x\to 0}\left(\frac {e^x}{x}-\frac {1}{\arctan{x}}\right)$ without using L'Hopital (with my attempts)

3 Answers3