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May 24th, 2017, 01:24 PM   #1
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Find limit

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May 24th, 2017, 01:49 PM   #2
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Don't you just need to divide through by $x$ to get a result out of that?
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May 25th, 2017, 10:17 PM   #3
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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!
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May 26th, 2017, 04:14 AM   #4
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Aye caramba! It's another sledgehammer-nut 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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