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I found a set of problems of limits which i can't seem to work my way around.

I tried using the natural log and then applying L'Hospital's rule but I can't seem to make it work. The problem was to find the limit of the function of $$\left(\frac{\tan x}{x}\right)^\frac1x$$ As x approaches 0 Please help me as there are some more problems like this. I cannot think of ways to evaluate them

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  • $\begingroup$ This handout of mine on the topic from the late 1990s may be of use. $\endgroup$ – Dave L. Renfro Nov 13 '18 at 14:33
  • $\begingroup$ What happened when you used logs? Where did you get stuck? Try to use standard limit $\lim\limits _{t\to 0}\dfrac{\log(1+t)}{t}=1$. $\endgroup$ – Paramanand Singh Nov 13 '18 at 18:21
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We have that (refer to this OP)

$$\lim_{x\to0}\frac{\tan x-x}{x^3}=\frac13 \implies \tan x=x+ \frac13x^3+o(x^3)$$

therefore

$$\left(\frac{\tan x}{x}\right)^\frac1x=\left(1+ \frac13x^2+o(x^2)\right)^\frac1x=\left[\left(1+ \frac13x^2+o(x^2)\right)^{\frac1{\frac13x^2+o(x^2)}}\right]^{\frac{\frac13x^2+o(x^2)}x}\to e^0=1$$

Frome the same derivation we can now easily derive that

$$\left(\frac{\tan x}{x}\right)^\frac1{x^2}\to \sqrt[3] e$$

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  • $\begingroup$ Could you guide me to a proof of t^(1/t)->1 as t->0 $\endgroup$ – user182947 Nov 13 '18 at 14:13
  • $\begingroup$ that's wrong, $t^{1/t}\to 0$ as $t\to 0^+$ $\endgroup$ – Pavel R. Nov 13 '18 at 14:17
  • $\begingroup$ @PavelR. Thanks I obtain teh correct result in a completely wrong way! Now I've fixed the derivation :) $\endgroup$ – gimusi Nov 13 '18 at 15:18
  • $\begingroup$ @user182947 Sorry for the confusion! Now it should be ok! $\endgroup$ – gimusi Nov 13 '18 at 15:19
  • $\begingroup$ How Did you bring in 'e'. Did you use the definition of e? $\endgroup$ – user182947 Nov 14 '18 at 1:25
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L'Hospital's rule is not the alpha and omega of limits computations!

To determine the limit of the logarithm, use Taylor's formula at order $3$ for the tangent: $$\frac{\tan x}x=\frac{x+\cfrac{x^3}3+o(x^3)}x=1+\frac{x^2}3+o(x^2),$$ so that $$\frac1x\log\Bigl(\frac{\tan x}x\Bigr)=\frac1x\log\Bigl(1+\frac{x^2}3+o(x^2)\Bigr)=\frac1x\Bigl(\frac{x^2}3+o(x^2)\Bigr)=\frac{x}3+o(x)\to 0$$ and finally $\;\biggl(\dfrac{\tan x}x\biggr)^{\!\tfrac 1x}$ tends to $1$ as $x$ tends to $0$.

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  • $\begingroup$ How did you remove the log? I didn't understand that $\endgroup$ – user182947 Nov 14 '18 at 0:23
  • $\begingroup$ I simply used Taylor's polynomial for $\log(1+u)$ near $u=0$: it is $u+o(u)$. $\endgroup$ – Bernard Nov 14 '18 at 0:43
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Recall that $$\lim_{x \to \infty} \left( 1 + \frac{k}{x}\right)^x = \lim_{x \to 0^+} \left(1 + kx\right)^{1/x} = e^k$$ and for any $\epsilon > 0$ there is some $\delta > 0$ such that $$0 < x < \delta \implies x < \tan x < x + \epsilon x^2.$$ Therefore, $$1 = \lim_{x \to 0^+} 1^x \leq \lim_{x \to 0^+} \left( \frac{\tan x}{x} \right)^{1/x} \leq \lim_{x \to 0^+} (1 + \epsilon x)^{1/x} = e^\epsilon.$$ Now take $\epsilon \to 0$.

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The limit of logarithm can be found without Taylor's formula:

$$ \begin{align*} \lim_{x\to 0}\frac{\log\left(\frac{\tan x}{x}\right)}{x}&= \underbrace{\lim_{x\to 0}\frac{\log\left(\frac{\tan x}{x}\right)}{\frac{\tan x}{x}-1}}_{=1}\cdot\lim_{x\to 0}\frac{\frac{\tan x}{x}-1}{x} =\lim_{x\to 0}\frac{\frac{\tan x}{x}-1}{x} =\lim_{x\to 0}\frac{\tan x-x}{x^2} \stackrel{\text{L'Hosp}}{=}\lim_{x\to 0}\frac{\frac{1}{\cos^2x}-1}{2x}\\[12pt] &=\lim_{x\to 0}\frac{1}{\cos^2x}\cdot\lim_{x\to 0}\frac{1-\cos^2x}{2x} =\lim_{x\to 0}\frac{\sin^2x}{2x} =\lim_{x\to 0}\frac{\sin x}{x}\cdot\lim_{x\to 0}\frac{\sin x}{2}=0 \end{align*} $$

The second limit is derived from the identity

$$ \lim_{t\to 1}\frac{\log t}{t-1}=1 $$

after substituion $t=\frac{\tan x}{x}$ as $x\to 0$.

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