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How would I go about solving this following limit?

$$\lim_{x\to 0} \frac{(1+\tan x)^{\frac{1}{x}}-e}{x}$$

My attempts:

Direct substitution yields the limit to be undefined, also ruling out the possibility of using L'Hospital's Rule.

I don't see any clever substitutions that can be made with this limit.

Would squeeze theorem help here? Maybe using the trig. identities:

$$-1 \le \cos x \le 1$$

and

$$-1 \le \cos x \le 1$$ EDIT

I attempted to break the limit down term by term.

So, for the first one:

$$y = \lim_{x\to 0} (1 + \tan x)^{1/x}$$

Taking the natural log of both sides:

$$\ln y = \lim_{x\to 0} \frac{\ln(1+\tan x)}{x}$$

Direct sub. yields $0/0$. Using L'Hospital's rule:

$$\ln y = \lim_{x\to 0} \frac{\frac{\sec^2{x}}{1+\tan x}}{1} = \frac{\sec^2{x}}{1+\tan x} = 1$$

Thus, $\ln y = 1$, so $y= e$

EDIT #2

Thanks to a random comment, it actually does help me:

$$\lim_{x\to 0} \frac{(1+\tan x)^{\frac{1}{x}}-e}{x}$$

$$\lim_{x\to 0} \frac{e-e}{0} = \frac{0}{0}$$

Thus, we can use L'Hospitals here:

$$\lim_{x\to 0} \frac{(\tan x+1)^{1/x} \left(\frac{\sec^2 x}{x(\tan(x)+1)}-\frac{\ln(\tan(x)+1))}{x^2}\right)}{1} = (\tan x+1)^{1/x} \left(\frac{\sec^2 x}{x(\tan(x)+1)}-\frac{\ln(\tan(x)+1))}{x^2}\right)$$

I haven't made any further progress, sadly.

Any help would be appreciated.

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  • $\begingroup$ In the last step: $\ln y = 1 \Rightarrow y = e$. $\endgroup$ – Simon S Jan 21 '15 at 23:48
  • $\begingroup$ @SimonS omg I'm an idiot. $\endgroup$ – Varun Iyer Jan 21 '15 at 23:50
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You may write, for $x$ near $0$, $$ \tan x=x+\frac{x^3}{3}+\mathcal{O}(x^5) $$ $$ \log(1+\tan x)=x-\frac{x^2}{2}+\mathcal{O}(x^3) $$ $$ \frac1x\log(1+\tan x)=1-\frac{x}{2}+\mathcal{O}(x^2) $$ then $$ e^{\frac1x\log(1+\tan x)}=e^{1-\frac{x}{2}+\mathcal{O}(x^2)}=e(1-\frac{x}{2}+\mathcal{O}(x^2)) $$ and $$ \frac{(1+\tan x)^{\frac{1}{x}}-e}{x}=\frac{e(1-\frac{x}{2}+\mathcal{O}(x^2))-e}{x}=-\frac{e}{2}+\mathcal{O}(x) $$ giving $\displaystyle -\frac{e}{2} $ as limit.

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  • $\begingroup$ So using taylor series instead of L'Hosptial's rule was a better approach? $\endgroup$ – Varun Iyer Jan 22 '15 at 0:17
  • $\begingroup$ @VarunIyer I think yes in this case, because the exponent $\frac1x$ doesn't really allow simplifications by derivation. Thanks. $\endgroup$ – Olivier Oloa Jan 22 '15 at 0:30
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You almost have the idea. I will just hint that instead of letting $y$ be $\lim_{x \to 0} (1+\tan x)^{1/x}$, let it be the expression that you want to get the limit of. That is, let $y = \frac{(1+\tan x)^{1/x} - e}{x}$. Then take the natural log before taking the limit.

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First observe asimtotic of $1+tan(x)$, it is 1+x+o(x). Than find asimtotic of $(1+x)^{1/x} $= $e^{{\frac{1}{x}}{ln(1+x)}}$=$e^{1/x}$= $e^{\frac{1}{x}*(x-x^2/2+o(x))}$, and that is like $e*e^{-x/2}$

Then we have in nominator asimtotic like: $e*e^{-x/2}-e=e*(e^{-x/2}-1)=e*(1-x/2+o(x)-1)=e*(-x/2)$ and your linit is like $ \lim_{x->0}\frac{e*(-x/2)}{x}=-e/2$

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i will use the maclaurin series $\ln(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3}+\cdots, \ \tan x = x + \frac{x^3}{3} + \cdots.$ now we can expand

$\begin{align} \ln(1 + \tan x) &= \tan x - \frac{\tan^2 x}{2} +\frac{\tan^3 x}{3} + \cdots \\ &= x + \frac{x^3}{3}+\cdots -\frac{x^2}{2}+\cdots + \frac{x^3}{3} + \cdots\\ &= x - \frac{x^2}{2} + \cdots \end{align}$

therefore $\frac{1}{x} \ln(1 + \tan x) = 1 - \frac{x}{2} + \cdots$ exponentiating the last result gives $$(1 + \tan x)^{1/x} = ee^{- x/2 + \cdots} = e\{1- x/2 + \cdots \}$$

finally, $$\lim_{x \to 0}\frac{(1 + \tan x)^{1/x} - e}{x} = -\frac{e}{2} $$

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One more way: rewrite the limit as (take $g(x) = (1+\tan x)^{\frac{1}{x}}$ $$ \lim_{x \to 0} \lim_{\epsilon \to 0} \frac{(1+\tan(x+\epsilon))^{\frac{1}{x+\epsilon}} - (1+\tan \epsilon)^\frac{1}{\epsilon}}{x} = \lim_{\epsilon \to 0}\lim_{x \to 0}\frac{(1+\tan(\epsilon +x))^{\frac{1}{\epsilon +x}} - (1+\tan \epsilon)^\frac{1}{\epsilon}}{x}\\ =\lim_{\epsilon \to 0}g'(\epsilon) $$ You can do it since $g(x)$ is a continuous function. Now take the Taylor series expansion of $\tan \epsilon$ and you will get the result.

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    $\begingroup$ Is continuity enough, or do you need uniform continuity? $\endgroup$ – GFauxPas Jan 22 '15 at 0:51
  • $\begingroup$ OK I admit to be precise I need to prove that $g'(0)$ exists. It seems to though: wolframalpha.com/input/?i=%281%2Btan%28x%29%29^%281%2Fx%29 $\endgroup$ – Alex Jan 22 '15 at 1:04
  • $\begingroup$ I'm not sure how you could argue $g'(0)$ existed without calculating it, and once you calculate $g'(0)$, you've already solved the problem. (I'm prepared to be proven wrong, though :)) $\endgroup$ – Strants Jan 22 '15 at 1:09
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Let's proceed in the following manner $$\begin{aligned}L &= \lim_{x \to 0}\frac{(1 + \tan x)^{1/x} - e}{x}\\ &= \lim_{x \to 0}\dfrac{\exp\left(\dfrac{\log(1 + \tan x)}{x}\right) - e}{x}\\ &= e\cdot\lim_{x \to 0}\dfrac{\exp\left(\dfrac{\log(1 + \tan x)}{x} - 1\right) - 1}{x}\\ &= e\cdot\lim_{x \to 0}\dfrac{\exp z - 1}{x}\text{ (putting }z = \dfrac{\log(1 + \tan x)}{x} - 1)\\ &= e\cdot\lim_{z \to 0}\dfrac{\exp z - 1}{z}\cdot\lim_{x \to 0}\frac{z}{x}\\ &= e\cdot 1\cdot\lim_{x \to 0}\frac{z}{x}\\ &= e\cdot\lim_{x \to 0}\frac{\log(1 + \tan x) - x}{x^{2}}\\ &= e\cdot\lim_{x \to 0}\left\{\frac{\log(1 + \tan x) - \tan x}{x^{2}} + \frac{\tan x - x}{x^{2}}\right\}\\ &= e\cdot\lim_{x \to 0}\frac{\log(1 + \tan x) - \tan x}{x^{2}} + e\cdot\lim_{x \to 0}\frac{\tan x - x}{x^{2}}\\ &= e\cdot\lim_{x \to 0}\frac{\log(1 + \tan x) - \tan x}{x^{2}} + e\cdot 0\\ &= e\cdot\lim_{x \to 0}\frac{\log(1 + \tan x) - \tan x}{\tan^{2}x}\cdot\frac{\tan^{2}x}{x^{2}}\\ &= e\cdot\lim_{x \to 0}\frac{\log(1 + \tan x) - \tan x}{\tan^{2}x}\cdot 1\\ &= e\cdot\lim_{t \to 0}\frac{\log(1 + t) - t}{t^{2}}\text{ (putting }t = \tan x)\\ &= e\cdot\lim_{t \to 0}\dfrac{\left(t - \dfrac{t^{2}}{2} + \cdots\right) - t}{t^{2}} = -\frac{e}{2}\end{aligned}$$ I have used the following limits $$\lim_{x \to 0}\frac{\log(1 + \tan x)}{x} = \lim_{x \to 0}\frac{\log(1 + \tan x)}{\tan x}\cdot \frac{\tan x}{x} = 1$$ so that $z \to 0$ $$\lim_{x \to 0}\frac{\tan x - x}{x^{2}} = 0$$ which can be proved using inequalities $\sin x < x < \tan x$ for $x \in (0, \pi/2)$ (although the proof is slightly tricky but easily available in MSE). Finally in the last step Taylor series for $\log (1 + t)$ is used.

Update: Proof for $$\lim_{x \to 0}\frac{\tan x - x}{x^{2}} = 0$$ Clearly we can consider only the case $x \to 0^{+}$ because the function under consideration is odd and the letting $x \to 0^{-}$ will only change the sign of the answer (which wont matter as the answer would come out as $0$). Now we can see that $$\begin{aligned}A &= \lim_{x \to 0^{+}}\frac{\tan x - x}{x^{2}}\\ &= \lim_{x \to 0^{+}}\frac{\sin x - x\cos x}{x^{2}\cos x}\\ &= \lim_{x \to 0^{+}}\frac{\sin x - x\cos x}{x^{2}\cdot 1}\\ &= \lim_{x \to 0^{+}}\frac{\sin x - x}{x^{2}} + \lim_{x \to 0^{+}}\frac{x - x\cos x}{x^{2}}\\ &= \lim_{x \to 0^{+}}\frac{\sin x - x}{x^{2}} + \lim_{x \to 0^{+}}x\cdot\frac{1 - \cos x}{x^{2}}\\ &= \lim_{x \to 0^{+}}\frac{\sin x - x}{x^{2}} + 0 \cdot\frac{1}{2}\\ &= \lim_{x \to 0^{+}}\frac{\sin x - x}{x^{2}}\\\end{aligned}$$ Next we have the inequality $$\sin x < x < \tan x = \frac{\sin x}{\cos x}$$ for $0 < x < \pi/2$ and hence $$\cos x < \frac{\sin x}{x} < 1$$ or $$\frac{\cos x - 1}{x} < \frac{\sin x - x}{x^{2}} < 0$$ Now taking limits as $x \to 0^{+}$ and noting that $$\frac{\cos x - 1}{x} = -2\cdot\frac{\sin^{2}(x/2)}{(x/2)^{2}}\cdot\frac{(x/2)^{2}}{x} \to 0$$ we get $$A = \lim_{x \to 0^{+}}\frac{\sin x - x}{x^{2}} = 0$$

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EDIT:

Here's a method which ends up involving very little computation. Notice that you've shown that the function

$$f(x) = \begin{cases}(1+\tan x)^\frac{1}{x} & x\not= 0\\ e & x=0 \end{cases}$$

is continuous, and that the limit you want to calculate is precisely $f'(0)$. If we define $g$ by

$$g(x) = \begin{cases}\frac{1}{x}\ln(1+\tan x) & x\not= 0 \\ 1 & x=0 \end{cases}$$ Then $f(x) = e^{g(x)}$, so if $g'(0)$ exists, we will have at once that $$f'(0) = e^{g(0)}g'(0)$$ by the chain rule. We have that

$$g'(0) = \lim_{x\to0} \frac{\frac{1}{x}\ln(1+\tan x) - 1}{x} = \lim_{x\to0} \frac{\ln(1+\tan x) - x}{x^2}$$

A single application of L'Hospital's rule yields

$$\lim_{x\to0} \frac{\frac{\sec^2x}{1+\tan x} - 1}{2x} = \lim_{x\to0} \frac{(1-\tan x) - 1}{2x}$$ or $$\lim_{x\to0} -\frac{\sin x}{2x\cos x}$$ which is easily seen to be $-\frac{1}{2}$. Thus, $f'(0)$ (the limit we wish to calculate) is $$f'(0) = -\frac{e}{2}$$


Here's a continuation of your use of L'Hospital's rule.

You have

$$\lim_{x\to 0} \frac{(1+\tan x)^\frac{1}{x} - e}{x} = \lim_{x\to0} (\tan x+1)^{1/x} \left(\frac{\sec^2 x}{x(\tan(x)+1)}-\frac{\ln(\tan(x)+1))}{x^2}\right)$$

Since you've already show that $\lim_{x\to0} (\tan x+1)^{1/x} = e$, we can clean this up to obtain

$$e\lim_{x\to0}\frac{x\sec^2x - (1+\tan x)\ln(1+\tan x)}{x^2(1+\tan x)} = e\lim_{x\to0}\frac{x-(\cos^2x+\sin x \cos x)\ln(1+\tan x)}{x^2(\cos^2x+\sin x \cos x)}$$

Since $\cos^2x + \sin x \cos x \to 1$ as $x\to0$, we may remove it from the denominator (based on the rules for limit arithmetic), yielding

$$e\lim_{x\to0}\frac{x-(\cos^2x+\sin x \cos x)\ln(1+\tan x)}{x^2}$$

We can then apply L'Hospital's rule again to get

$$e\lim_{x\to0} \frac{1 - \left[-2\cos x \sin x - \cos^2x + \sin^2x)\ln(1+\tan x) + (\cos^2 x + \sin x \cos x)\frac{\sec^2x}{1+\tan x}\right]}{2x}$$

After applying various trigonometric identities, this simplifies to

$$e\lim_{x\to0}\frac{(\sin 2x - \cos 2x)\ln(1+\tan x)}{2x}$$ or $$e\lim_{x\to0}\frac{\sin 2x\ln(1+\tan x)}{2x} - e\lim_{x\to0}\frac{\cos 2x\ln(1+\tan x)}{2x} $$

which, since $\cos 2x \to 0$ and $\frac{\sin 2x}{2x} \to 1$, is $$-e\lim_{x\to0}\frac{\ln(1+\tan x}{2x}$$

Which, after one final application of L'Hospital's rule, comes to

$$-e\lim_{x\to0}\frac{\sec^2x}{2(1+\tan x)} = -\frac{e}{2}$$

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