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December 2nd 2010, 01:57 AM
hmmmm

test

we have $\displaystyle\lim_{x\to\infty} ( \frac{16x}{16x+5})^{9x}$

re-writing this as $\displaystyle\lim_{x\to\infty}exp^[\ln(( \frac{16x}{16x+5})^{9x})]$

$\displaystyle\lim_{x\to\infty}exp^{9x}[\ln\( \frac{16x}{16x+5})]$

Then using the substitution $\frac{1}{t}$

we have $\displaystyle\lim_{t\to 0}exp^{\frac{9}{t}\ln\frac{\frac{16}{t}}{\frac{16} {t}+5}$

Then using L'hopitals rule on $\displaystyle\lim_{t\to 0}{\frac{9}{t}\ln\frac{\frac{16}{t}}{\frac{16}{t}+ 5}$ = $\displaystyle\lim_{t\to 0}\frac{-45}{16+5t}$ = $\frac{-45}{16}$

So we have that

$\displaystyle\lim_{x\to\infty} ( \frac{16x}{16x+5})^{9x}$
= $\displaystyle\exp^\frac{-45}{16}$
December 2nd 2010, 02:35 AM
CaptainBlack

Quote:

Originally Posted by hmmmm

If we write $\displaystyle( \frac{16x}{16x+5})^{9x}$ as $\displaystyle (1-\frac{5}{16x+5})^{9x}$

we then have $\displaystyle\lim_{x\to\infty} (1-\frac{5}{16x+5})^{9x}$

re-writing this as $\displaystyle\lim_{x\to\infty}exp^[\ln((1-\frac{5}{16x+5})^{9x})]$

$\displaystyle\lim_{x\to\infty}exp^{9x}[\ln((1-\frac{5}{16x+5}))]$

Then using the substitution $\frac{1}{t}$

we have $\displaystyle\lim_{t\to 0}exp^({\frac{9}{t}[\ln(1-\frac{5}{16(\frac{1}{t})+5})])$

Then using L'hopitals rule on $\displaystyle\lim_{t\to 0}{\frac{9}{t}[\ln(1-\frac{5}{16(\frac{1}{t})+5})])$

$\displaystyle \lim_{x\to \infty}(\frac{16x}{16x+5})^{9x})=e^\frac{-45}{16}$

Try elastic brackets \left( and \right) they expand to the size needed for what is enclosed, same for \left[ , \right] and \left\{, \right\}

CB
December 2nd 2010, 08:06 AM
Soroban

Hello, hmmmm!

Use larger parentheses/brackets as CaptainBlack suggested.

And here is another approach to the limit . . .

We have $\displaystyle\lim_{x\to\infty} \left(\frac{16x}{16x+5}\right)^{9x}$

Ignore the limit for now . . .

$\text{Divide top and bottom by }16x\!:\;\;\dfrac{1}{1 + \frac{5}{16x}} \;=\;\dfrac{1}{1 + \frac{1}{\frac{16}{5}x}}}$

$\text{Then: }\;\left(\dfrac{1}{1 + \frac{1}{\frac{16}{5}x}}}\right)^{9x} \;=\; \left[\left(\dfrac{1}{1 + \frac{1}{\frac{16}{5}x}}}\right)^{\frac{16}{5}x}\r ight]^{\frac{45}{16}} \;=\; \left[\dfrac{1}{\left(1 + \frac{1}{\frac{16}{5}x}\right)^{\frac{16}{5}x}}\ri ght]^{\frac{45}{16}}$

$\displaystyle \text{Hence: }\;\lim_{x\to\infty} \left[\dfrac{1}{\left(1 + \frac{1}{\frac{16}{5}x}\right)^{\frac{16}{5}x}}\ri ght]^{\frac{45}{16}} \;=\; \left[\dfrac{1}{\lim \left(1 + \frac{1}{\frac{16}{5}x}\right)^{\frac{16}{5}x}}\ri ght]^{\frac{45}{16}}$ .[1]

$\displaystyle \text{We note that: }\:\lim_{x \to \infty} \left(1 + \frac{1}{\frac{16}{5}x}}\right)^{\frac{16}{5}x} \;=\;e$

Therefore, [1] becomes: . $\displaystyle \left(\frac{1}{e}\right)^{\frac{45}{16}} \;=\;e^{-\frac{45}{16}}$