Given this limit:

$$\lim_{x\to\infty} [\log_{10}{(5\cdot6^x+2\sqrt{x^3}})-\log_{10}{(\sin{x}+4\cdot6^x+\cos{x})]}$$

I simplified:

$$\lim_{x\to\infty} \log_{10}{\frac{5\cdot6^x+2\sqrt{x^3}}{\sin{x}+4\cdot6^x+\cos{x}}}$$

Now, I can't understand why I can cancel $6^x$, $2\sqrt{x^3}$, $\sin{x}$, $6^x$ and $\cos{x}$ so that the solution is $\log_{10}{\frac{5}{4}}$ as my professor told me.

  • $\begingroup$ For large x the terms with $6^x$ completely dominate the others, we need only to look at $log_{10}(5\times 6^x)-log_{10}(4\times 6^x)=log_{10}(\frac{5}{4})$. $\endgroup$ – herb steinberg May 3 '18 at 0:26

$$\lim_{x\to\infty} \log_{10}{\frac{5\cdot6^x+2\sqrt{x^3}}{\sin{x}+4\cdot6^x+\cos{x}}}= \lim_{x \to \infty}\log_{10}{\frac{5+2\frac{\sqrt{x^3}}{6^x}}{\frac{\sin{x}}{6^x}+4+\frac{\cos{x}}{6^x}}}=\log_{10}\frac54$$ Edit: \begin{align} 0 \le \lim_{x \to \infty}\frac{\sqrt{x^3}}{6^x} &\le \lim_{x \to \infty}\frac{x^2}{e^{(\log 6) x}} \le \lim_{x \to \infty}\frac{x^2}{e^x} \le \lim_{x \to \infty}\frac{x^2}{1+x+\frac{x^2}{2}+\frac{x^3}{6}} = 0 \end{align}

  • $\begingroup$ What does this mean? $\endgroup$ – user557276 May 2 '18 at 18:54
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    $\begingroup$ Can you check that $\lim_{x \to \infty }\frac{\sqrt{x^3}}{6^x}=0$? $\endgroup$ – Siong Thye Goh May 2 '18 at 18:55
  • $\begingroup$ Isn't it infinity/infinity? $\endgroup$ – user557276 May 2 '18 at 18:57
  • $\begingroup$ if you substitute $\infty$ inside directly, you do observe $\frac{\infty}{\infty}$, we call this indeterminate form and you might like to check out tricks like L'hopital rule $\endgroup$ – Siong Thye Goh May 2 '18 at 18:59
  • $\begingroup$ I know about L'Hopital rule but I can't use it in my exam. $\endgroup$ – user557276 May 2 '18 at 19:01

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