I have difficulty to conclude this limit ....; place of my attempts and results, can anyone help?

tanks in advance

$$\lim_{x\to +a}\, \left(1+6\left(\frac{\sin x}{x^2}\right)^x\cdot\frac{\log(1+10^x)}{x}\right),\quad a=+\infty,\,\,\,0,\,\,\,\,-\infty$$ 1):$\,\,\,{a=+\infty}$

$$\lim_{x\to +\infty}\left(1+6\left(\frac{\sin x}{x^2}\right)^x\cdot\frac{\log(1+10^x)}{x}\right)$$ $$\sim\lim_{x\to +\infty}\, 1+6\left( \frac{\sin^x x}{x^{2x}} \cdot\frac{\log 10^x }{x}\right)=\lim_{x\to +\infty}\, 1+6\left( \frac{\sin^x x}{x^{2x}} \right)$$

$$\text{observing}\,\,\,\,\,\left|\frac{\sin x}{x^2} \right|<\frac{1}{x^2}, \text{so} \left|\left(\frac{\sin x}{x^2}\right)^x\right|<\frac{1}{x^{2x}}\to 0\,\,\,\, x\to+\infty:$$ $$\text{infact we have} \lim_{x\to+\infty} \frac{1}{x^{2x}}=\lim_{x\to +\infty} e^{-2x\log x}=0$$ $$=\lim_{x\to +\infty}\, 1+6\left( \frac{\sin^x x}{x^{2x}} \right)=1+6\cdot0=1 $$ 2):$\,\,\,{a=0}$

$$\lim_{x\to 0}\, \left(1+6\left(\frac{\sin x}{x^2}\right)^x\cdot\frac{\log(1+10^x)}{x}\right) \sim\lim_{x\to 0}\, 1+\frac{6\log2}{x}\left(\frac{ x}{x^2}\right)^x$$ $$\sim\lim_{x\to0}1+\frac{6\log2}{x}\left(\frac{1}{x}\right)^x$$ $$=\lim_{x\to 0} 1+\frac{6\log2}{x}e^{ x\ln \left(\frac{1}{x}\right)}\to \lim_{x\to 0}\, x\ln \left(\frac{1}{x}\right)=\lim_{x\to 0}\, \frac{\ln \left(\frac{1}{x }\right)}{\frac{1}{x}}=0\to e^0=1$$ $$=\lim_{x\to 0}\,1+\frac{6\log2}{x}\cdot1=+\infty$$


let $x=-t,\,\,\,$if$\,\,\,\ x\to -\infty\,\,\,$we have$ \,\,\,\ t\to+\infty,\,\,\,$ and so : $$\lim_{t\to +\infty}\, \left(1+6\left(\frac{\sin(-t)}{t^2}\right)^{-t}\cdot\frac{\log(1+10^{-t})}{-t}\right)$$ $$=\lim_{t\to +\infty}\, \left(1-6\left(-\frac{\sin t }{t^2}\right)^{-t}\cdot\frac{\log\left(1+\frac{1}{10^t}\right)}{t}\right)$$ $$=1-6\cdot\lim_{t\to +\infty}\, \left[\left(-\frac{t^2}{\sin t }\right)^{t}\cdot\frac{\log\left(1+\frac{1}{10^t}\right)}{t}\right]$$ $$\stackrel{(\bf T)}{=}1-6\cdot\lim_{t\to +\infty}\, \left[\left(-\frac{t^2}{\sin t }\right)^{t}\cdot \left(\frac{1}{10^t}-\frac{1}{2\cdot10^{2t}}\cdot\frac{1}{t}\right) \right]$$

.... but here i'm lost ....

  • 4
    $\begingroup$ I think you may have some issues with the term $({\sin x\over x^2})^x$; this isn't always real-valued. $\endgroup$ – David Mitra Mar 28 '12 at 15:32

As pointed by David Mitra in comments, the function $$f(x) = \left\{1 + 6\left(\frac{\sin x}{x^{2}}\right)^{x}\frac{\log(1 + 10^{x})}{x}\right\}$$ is not well defined whenever $x \to \pm \infty$ because $\sin x$ becomes negative. The same holds when $x \to 0$. However if we restrict ourselves to $x \to 0^{+}$ then the function is well defined near $0$ and hence we may try to evaluate its limit as $x \to 0^{+}$. Clearly we can see that when $x \to 0^{+}$ then $\log(1 + 10^{x}) \to \log 2$ so that we ideally need to take care of the part $$g(x) = \frac{(\sin x)^{x}}{x^{2x + 1}}$$ which is better handled by taking logarithm. We have $$\begin{aligned}\log g(x) &= x\log\sin x - (2x + 1)\log x\\ &= x\log\left(\frac{\sin x}{x}\right) + x\log x - 2x\log x - \log x\\ &= x\log\left(\frac{\sin x}{x}\right) - x\log x - \log x\end{aligned}$$ Now as $x \to 0^{+}$ we can see that $(\sin x)/x \to 1$ so that first term tends to $0$. The second term $x\log x$ also tends to $0$ and the last term $-\log(x)$ tends to $\infty$. So that $\log g(x)$ tends to $\infty$ as $x \to 0^{+}$ and hence $g(x)$ also tends to $\infty$ as $x \to 0^{+}$. Since $f(x) = 1 + 6g(x)\log(1 + 10^{x})$ it follows that $f(x) \to \infty$ as $x \to 0^{+}$.


$\lim_{x\to +a}\, \left(1+6\left(\frac{\sin x}{x^2}\right)^x\cdot\frac{\log(1+10^x)}{x}\right),\quad a=+\infty,\,\,\,0,\,\,\,\,-\infty$

When $a=-\infty$,prove as below

take $x=2k\pi-\frac{\pi}{2}$,($k<0,k\in Z$) ,$(\frac{\sin x}{x^2})^x=(\frac{-1}{(2k\pi-\frac{\pi}{2})^2})^{2k\pi-\frac{\pi}{2}}$,when $k \to -\infty$, $-1^{2k\pi-\frac{\pi}{2}}$ may not real number.so there is no limit ,when $x \to -\infty$.

  • 5
    $\begingroup$ The exponent of $|{\sin x\over x^2}|^x$ is negative; it is not bounded. $\endgroup$ – David Mitra Mar 28 '12 at 16:43

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