$$\lim_{x\to 0}\frac{\sqrt[5]{1+\sin(x)}-1}{\ln(1+\tan(x))}$$
How to evaluate the limit of this function without using L'Hôpital's rule?
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Hint: Use the binomial series for the numerator, and the Mercator series for the denominator. – Lucian Oct 07 '15 at 17:09
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Or: multiply and divide by $\tan x$. Recall the limit of $\frac{\ln(1+x)}{x}$, and recall the definition of a derivative. – Noam Shalev - nospoon Oct 07 '15 at 17:14
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I got $$\lim_{x\to0}\frac{(1+x)^\frac{1}{5}-1}{tgx}$$ what now? – Nick202 Oct 07 '15 at 17:49
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You mean $\lim_{x\to 0}\frac{(1+\sin x)^\frac{1}{5}-1}{tgx}$. now, $\frac{1}{\tan x}=\cos x \frac{1}{\sin x}$. the cosine goes to 1, and for the sine make a substitution. what is, by definition, the derivative of $x^n$ at 1? – Noam Shalev - nospoon Oct 07 '15 at 19:00
3 Answers
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Use Taylor's developments at order $1$ and compose them: \begin{align*}&\left.\begin{array}{l} \sin x=x+o(x)\\\sqrt[5]{1+u}=1+ \dfrac15u+o(u) \end{array}\right\}\Rightarrow\enspace\sqrt[5]{1+\sin x}-1= 1+ \frac15x+o(x)-1=\frac15x+o(x)\\[1ex] &\left.\begin{array}{l} \tan x=x+o(x)\\\ln(1+u)=u+o(u) \end{array}\right\}\Rightarrow\enspace\ln{1+\tan x}= x+o(x) \end{align*} whence $$\frac{\sqrt[5]{1+\sin x}-1}{\ln(1+\tan x)}=\frac{\dfrac15x+o(x)}{x+o(x)}=\frac{\dfrac15+o(1)}{1+o(1)}\to \frac15.$$
Bernard
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This is the approach that I would use. But, I thought that I would post another way forward that circumvents the use of L'Hospital's Rule, asymptotic analysis, or other derivative-based methodologies. Instead, I relied on just basic inequalities and the squeeze theorem. I would like to hear your thoughts if you have time. – Mark Viola Oct 07 '15 at 18:25
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I thought that it would be instructive to present a way forward that circumvents the use of L'Hospital's Rule, asymptotic analysis, or other derivative-based methodologies. To that end, herein, we use some basic inequalities and the squeeze theorem.
In THIS ANSWER for $x>-1$, I showed
$$\frac{x}{x+1}\le\log (1+x)\le x \tag 2$$
using only Bernouli's Inequality and the limit definition of the exponential function.
And here, the inequality
$$|x\cos x|\le |\sin x|\le |x| \tag 1$$
was established by appealing to geometry only.
Using $(1)$ and $(2)$, we have for $x>0$
$$\frac{(1+x\cos x)^{1/5}-1}{\tan x}\le\frac{(1+\sin x)^{1/5}-1}{\log (1+\tan x)}\le\frac{(1+x)^{1/5}-1}{\frac{\tan x}{1+\tan x}} \tag 3$$
Note that the right-hand side 0f $(3)$ can be written
$$\begin{align} \frac{(1+x)^{1/5}-1}{\frac{\tan x}{1+\tan x}}& =(1+\tan x)\left(\frac{x\cos x}{\sin x}\right)\\\\\ &\times \left(\frac{1}{1+(1+x)^{1/5}+(1+x)^{2/5}+(1+x)^{3/5}+(1+x)^{4/5}}\right)\\\\ &\to \frac 15\,\,\text{as}\,\,x\to 0 \end{align}$$
Similarly, the left-hand side 0f $(3)$ can be written
$$\begin{align} \frac{(1+x\cos x)^{1/5}-1}{\tan x}& =(\cos^2 x)\left(\frac{x}{\sin x}\right)\\\\ &\times \left(\frac{1}{1+(1+x\cos x)^{1/5}+(1+x\cos x)^{2/5}+(1+x\cos x)^{3/5}+(1+x\cos x)^{4/5}}\right)\\\\ &\to \frac 15\,\,\text{as}\,\,x\to 0 \end{align}$$
Therefore, by the squeeze theorem
$$\lim_{x\to 0^+}\frac{(1+\sin x)^{1/5}-1}{\log (1+\tan x)}=\frac15$$
A similar development for $x<0$ results in the same limit. Therefore, we have
$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to 0}\frac{(1+\sin x)^{1/5}-1}{\log (1+\tan x)}=\frac15}$$
and we are done without using anything other than standard inequalities!
Mark Viola
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That's tricky! I wonder if one can show $\mathrm e^x$ is convex directly without derivatives from the limit definition. Proving $;\mathrm e^{\frac{x+y}2}\le \dfrac12(\mathrm e^x+\mathrm e^y)$ would be enough. – Bernard Oct 07 '15 at 19:15
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@Bernard Perhaps, but it isn't clear how the concavity alone of the log function is sufficient to provide the bounds that permit the "squeeze" of the limit of interest herein. – Mark Viola Oct 07 '15 at 19:40
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I'm not sure if this can be made formal...
$$\lim_{x\to 0}\frac{\sqrt[5]{1+\sin(x)}-1}{\ln(1+\tan(x))}$$
$$\sqrt[5]{1+x} = 1 + \dfrac x5 - \dfrac{2}{25}x^2 + O(x^3)$$
$$\sqrt[5]{1+\sin(x)}-1 \approx \dfrac 15 \sin x - \dfrac{2}{25}\sin{x^2}$$
$$\ln(1+x) = x - \dfrac 12x^2 + O(x^3)$$
$$\ln(1+ \tan x) \approx \tan x - \dfrac 12 \tan^2 x $$
\begin{align} \frac{\sqrt[5]{1+\sin(x)}-1}{\ln(1+\tan(x))} & \approx \frac{\dfrac 15 \sin x-\dfrac{2}{25}\sin{x^2}}{\tan x-\dfrac 12 \tan^2 x} \\ & \approx \frac{\dfrac 15 \sin x-\dfrac{2}{25}\sin{x^2}}{\tan x-\dfrac 12 \tan^2 x} \\ & \approx \dfrac{\sin x}{\tan x} \frac{\dfrac 15 - \dfrac{2}{25}\sin x}{1 - \dfrac 12 \tan x} \\ & \approx \cos x \, \frac{\dfrac 15 - \dfrac{2}{25}\sin x}{1 - \dfrac 12 \tan x} \\ & \to \dfrac 15 \quad \text{as} \quad x \to 0 \end{align}
Steven Alexis Gregory
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