4

I tried replacing $x$ with $0$ the log returns $1$ and the $1/\sin x$ returns $1/0$. So I thought the limit should be infinity. However, graphing the function yields undefined value at $0$, and the result shows that the limit is $e^{2\,e^ {- 1 }}$. I have no idea how did they reach that conclusion.

Anton Vrdoljak
  • 2,717
randomtrekker
  • 57
  • 1
    The discrepancy comes from the way how your graphing tool interprets your expression: https://www.desmos.com/calculator/vp3gmcoc0g – trancelocation Jul 29 '20 at 11:14
  • 2
    If teh exponent $;\frac1{\sin x};$ is only on the argument $;(e+2x);$, then you're right: the limit doesn't exist. Yet, by what you say the limit is, I suspect the exponent is on the whole logarithm, meaning: your expression is $$\left(\log(e+2x)\right)^{\frac1{\sin x}}$$ – DonAntonio Jul 29 '20 at 11:20

3 Answers3

5

Using Taylor polynomials \begin{align*} \log (\mathrm{e} + 2x)^{\frac{1}{{\sin x}}} & = \exp \left( {\frac{{\log \log (\mathrm{e} + 2x)}}{{\sin x}}} \right) = \exp \left( {\frac{{\log \left( {1 + \log \left( {1 + \frac{{2x}}{\mathrm{e}}} \right)} \right)}}{{\sin x}}} \right) \\ & = \exp \left( {\frac{{\log \left( {1 + \frac{{2x}}{\mathrm{e}} + \mathcal{O}(x^2 )} \right)}}{{\sin x}}} \right) = \exp \left( {\frac{{\frac{{2x}}{\mathrm{e}} + \mathcal{O}(x^2 )}}{{\sin x}}} \right) \\ & = \exp \left( {\frac{2}{\mathrm{e}}\frac{x}{{\sin x}}(1 + \mathcal{O}(x))} \right). \end{align*} Thus, the limit is indeed $\mathrm{e}^{2/\mathrm{e}}$.

Gary
  • 31,845
3

$$\lim_{x\to0}\left(\ln(e+2x)\right)^{1/\sin x}$$

$$=\lim_{x\to0}\left(1+\ln\left(1+\dfrac{2x}e\right)\right)^{1/\sin x}$$

$$=\left(\lim_{x\to0}\left(1+\ln\left(1+\dfrac{2x}e\right)\right)^{\dfrac1{\ln\left(1+\dfrac{2x}e\right)}}\right)^{\lim_{x\to0}\dfrac{\ln\left(1+\dfrac{2x}e\right)}{\sin x}}$$

The inner limit converges to $e$ as $\lim_{n\to\infty}\left(1+\dfrac1n\right)^n=e$

Now for the exponent

$$\lim_{x\to0}\dfrac{\ln\left(1+\dfrac{2x}e\right)}{\sin x}=\dfrac 2e\cdot\lim_{x\to0}\dfrac{\ln\left(1+\dfrac{2x}e\right)}{\dfrac{2x}e}\cdot\lim_{x\to0}\dfrac x{\sin x}=?$$

lab bhattacharjee
  • 274,582
1

We have $ \ln (e+2x)^{1/ \sin x}= e^{\frac{\ln (e+2x)}{\sin x}}.$

Now compute $ \lim_{x \to 0}\frac{\ln (e+2x)}{\sin x}$ with l'Hospital.

Fred
  • 77,394