# Find $\lim_{x\to 0}(x+\sin x)^{\tan x}$

Find the limit $\lim_{x\to 0}(x+\sin x)^{\tan x}.$

I tried:

$\lim_{x\to 0}{\tan x}.\ln(x+\sin x)=\lim_{x\to 0}\frac{\ln(x+\sin x)}{\cot x}=\lim_{x\to 0}\frac{(1+\cos x)(\cos^2 x)}{-(x+\sin x)}$

I stuck at this step.

• Hint: $${\tan x}.\ln(x+\sin x) = \tan x \left(\ln x + \ln \left(1+\frac{\sin x}{x}\right)\right) = \tan x \ln x + \tan x \ln \left(1+\frac{\sin x}{x}\right)$$ – r9m Jan 8 '15 at 11:59
• maybe L'Hopital's rule? – Irrational Person Jan 8 '15 at 12:02
• Derivative of cotangent is $-1/\sin^2 x$, not $\cos$ – sas Jan 8 '15 at 12:03
• Please do not use titles consisting only of math expressions; these are discouraged for technical reasons -- see meta. – Martin Sleziak Jan 8 '15 at 12:10

You did a mistake in the last identity, it should be $$\lim_{x\to 0}\frac{\ln(x+\sin x)}{\cot x}=\lim_{x\to 0}\frac{(1+\cos x)(-\sin^2 x)}{-(x+\sin x)}$$ And after that another L'Hopital gives $$\lim_{x\to 0}\frac{(1+\cos x)(-\sin^2 x)}{-(x+\sin x)}=\lim_{x\to 0}\frac{-\sin^3x + (1+\cos x)2\sin x\cos x}{1+\cos x}=0$$

• yay more L'Hopital's rule! – Irrational Person Jan 8 '15 at 12:12
• L'Hopital is not important here, the goal is to point out the mistake in order to OP could finish his own solution. – sas Jan 8 '15 at 12:18

Hints:

$$\left(x+\sin x)\right)^{\tan x}=e^{\tan x\log(x+\sin x)}$$

And now, using l'Hospital's rule"

$$\lim_{x\to 0}\frac{\log(x+\sin x)}{\cot x}\stackrel{\text{l'H}}=\lim_{x\to 0}\frac{\frac{1+\cos x}{x+\sin x}}{-\frac1{\sin^2x}}$$

And

$$\lim_{x\to 0}\frac{\sin^2x}{x+\sin x}\stackrel{\text{l'H}}=\lim_{x\to 0}\frac{2\sin x\cos x}{1+\cos x}=0$$

Deduce now that the limit is $\;1\;$ using continuity of the exponential function

• yay, L'Hopital's rule – Irrational Person Jan 8 '15 at 12:05
• @Bot Indeed so. Love this useful rule. – Timbuc Jan 8 '15 at 15:27

It's simpler with equivalents: $$\tan x \sim_0 x,\quad\ln(x+\sin x)=\ln x + \ln\Bigl(1+\dfrac{\sin x}x\Bigr)\sim_0 \ln x$$ hence $\,\,\tan x\ln(x+\sin x)\sim_0 x\ln x$ so that $\,\,\tan x\ln(x+\sin x)\underset{x\to 0}{\longrightarrow}0\,\,$ and finally $$(x+\sin x)^{\tan x}\underset{x\to 0}{\longrightarrow} 1.$$

$$\lim_{x\to0}{\tan x\ln(x+\sin x)}={\lim_{x\to0}\dfrac{\ln(x+\sin x)}{\cot x}}\stackrel{\text{L'H}}={\lim_{x\to0}\dfrac{\dfrac{1+\cos x}{x+\sin x}}{-\dfrac1{\sin^2x}}}=\lim_{x\to0} \left( -\dfrac{\sin^2x(1+\cos x)}{x+\sin x} \right) \stackrel{\text{L'H}}=\lim_{x\to0}{\dfrac{\sin^3x-2\sin x\cos x(\cos x+1)}{\cos x+1}}={\lim_{x\to0}\sin x(1-3\cos x)}=0$$

• It'd be probably a very good idea not to make that "fat" the expression in the exponent of $\;e\;$ . Maybe it is better to calculate the limit separatedly or else use $\;\exp\;$ instead. – Timbuc Jan 8 '15 at 12:13
• @Timbuc. Better? – user164524 Jan 8 '15 at 12:20
• Much better, imo...yet I can't see how you got the last expression: it doesn't look what you'd get after doing l'Hospital in the rightmost expression in the first line. – Timbuc Jan 8 '15 at 15:29
• @Timbuc. I just applied L'Hopital's rule and simplified at once because I thought that it is obvious what I did. But, OK, I edited question. – user164524 Jan 9 '15 at 10:08

$$(x+\sin(x))^{\tan(x)}=e^{\tan(x)\ln(x+\sin(x))}$$

Compute $$\lim_{x\to0}\frac{\ln(x+\sin(x))}{\frac{1}{\sin(x)}}=\lim_{x\to0}\frac{\frac{1+\cos(x)}{x+\sin(x)}}{\frac{-\cos(x)}{\sin^2(x)}}=-2\lim_{x\to0}\frac{\sin(x)}{\frac{x}{\sin(x)}+1}=0$$

And use above.

• Why $\frac{1}{\text{sin}(x)}$ in denominator? – Alex Silva Jan 8 '15 at 12:04
• @AlexSilva Why not? – Pp.. Jan 8 '15 at 12:06
• @AlexSilva Calm down. It is just a question. – Pp.. Jan 8 '15 at 12:08
• @AlexSilva That way you turn the leftmost expression in that second line of mathematical expressions into an indeterminate $\;\frac\infty\infty\;$ and you can then apply l'Hospital. Perhaps Pp. forgot to remark l'Hospital...or perhaps he left that for the OP to complete. – Timbuc Jan 8 '15 at 12:11
• @Timbuc, I know that, but why $\frac{1}{\text{sin}(x)}$ if the real denominator is $\frac{1}{\text{tan}(x)}$? This was my question. – Alex Silva Jan 8 '15 at 12:14

This is merely a warning and a tip of intuition than an answer. It became too long for just being a comment. If this is considered bad habit, please tell me, and I'll remove it.

Firstly a comment on the existing solutions. I think that one should be careful with $\ln(x+\sin x)$, since $x+\sin x<0$ if $x<0$ and $\ln$ is only defined for positive values.

Since the question have tag "real-analysis", and the expression $(x+\sin x)^{\tan x}$ is complex for (most) negative $x$, I therefore suggest that one should calculate the one-sided limit $$\lim_{t\to 0^+}(x+\sin x)^{\tan x}$$ only.

There are already many correct answers doing this (if one assumes $x>0$ in the solutions), so I won't write yet another one, but let me just point out that before going into too complicated calculations, we can use intuition:

When $x$ is small $\sin x\approx x$ and $\tan x\approx x$, so one would expect the limit to be the same as the one of $(x+x)^x$. But $$(2x)^x=e^{x\ln(2x)}.$$ Since $x\ln(2x)\to 0$ as $x\to0^+$ and the exponential function is continuous at $0$ (with value 1), the limit should have value $1$. I emphasize that this is not a proof, but it could be made into a proof by expanding $\sin x$ and $\tan x$ and keeping the error term.

Just as r9m commented, take logarithms of both sides. So, starting with $$A=\big(x+\sin (x)\big)^{\tan (x)}$$ $$\log(A)=\tan(x) \log\big(x+\sin (x)\big)=\tan(x) \Big(\log(x)+\log\big(1+\frac{\sin (x)}{x}\big)\Big)$$ Now, remember that, close to $x=0$, $\tan(x)=x+O\left(x^2\right)$. The second logarithm tends to $\log(2)$; so $$\log(A)\approx x\big(\log(x)+\log(2)\big)=x\log(x)+x\log(2)$$ I am sure that you can finish from here.