May 24th, 2017, 01:24 PM  #1 
Newbie Joined: Oct 2015 From: London Posts: 20 Thanks: 0  Find limit 
May 24th, 2017, 01:49 PM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,941 Thanks: 2267 Math Focus: Mainly analysis and algebra 
Don't you just need to divide through by $x$ to get a result out of that?

May 25th, 2017, 10:17 PM  #3 
Newbie Joined: Dec 2016 From: Austin Posts: 11 Thanks: 1  Limit at Infinity
You can utilize L'Hospital's Rule. Suppose: $\displaystyle \lim_{x \rightarrow \infty} \frac{f(x)}{g(x)}$ yields $\displaystyle 0$ or $\displaystyle \pm \infty$ in both the numerator and denominator. If: $\displaystyle \lim_{x \rightarrow \infty} \frac{f'(x)}{g'(x)}$ yields a finite value or $\displaystyle \pm \infty$, then $\displaystyle \lim_{x \rightarrow \infty} \frac{f(x)}{g(x)} = \lim_{x \rightarrow \infty} \frac{f'(x)}{g'(x)} $ Begin by taking the derivative of the numerator and denominator: $\displaystyle \lim_{x \rightarrow \infty} \frac{1}{1+ \frac{2 \log (x)}{x \ln(10)}}$ It may be hard to evaluate the integral in this form, so consider simplifying the complex fraction. Start by simplifying the denominator: $\displaystyle \lim_{x \rightarrow \infty} \frac{1}{\frac{x \ln(10)+ 2 \log (x)}{x \ln(10)}}$ Flip the fraction: $\displaystyle \lim_{x \rightarrow \infty} \frac{x \ln(10)}{x \ln(a)+ 2 \log (x)}$ Because this still yields an indeterminate form, repeat L'Hospital's Rule once again, by taking the derivative of the numerator and denominator: $\displaystyle \lim_{x \rightarrow \infty} \frac{ \ln(10) }{ \ln(10)+ \frac{2}{x \ln(10)}}$ This limit can indeed be evaluated, if we consider the following axiom: $\displaystyle \lim_{x \rightarrow \infty} \frac{a}{bx}=0$ for any constants $\displaystyle a,b \epsilon \mathbb{R} $ and $\displaystyle b \neq 0$ Applying this to our limit yields: $\displaystyle \lim_{x \rightarrow \infty} \frac{ \ln(10) }{ \ln(10)+ 0}$ = 1 Thus, our answer is $\displaystyle 1$. Hope that helps! 
May 26th, 2017, 04:14 AM  #4 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,941 Thanks: 2267 Math Focus: Mainly analysis and algebra 
Aye caramba! It's another sledgehammernut moment. I think it's better to notice that $$\lim_{x \to \infty} \frac{\log^a x}{x^b}=0 \quad \text{for all $a \gt 0, \, b \gt 0$}$$ 

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