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$\\displaystyle{\\lim_{n \\to \\infty}\\left(\\frac{1}{n^{1/\\var{r}}}\\right)=} $ [[0]]

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$\\displaystyle{\\lim_{n \\to \\infty}\\left(\\var{k1}^{1/n}\\right)=} $ [[0]]

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$\\displaystyle{\\lim_{n \\to \\infty}\\left(\\var{k}^{1/n}\\right)=} $ [[0]]

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$\\displaystyle \\lim_{n \\to \\infty}\\left(\\simplify[std]{({c}n+{d})/({al}n-{ga})}\\right) = $ [[0]]

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Enter your answer as a fraction or integer, not as a decimal.

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$\\displaystyle{\\lim_{n \\to \\infty}\\left(\\simplify[std]{{c}/{n}}\\right)^n=} $ [[0]]

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$\\displaystyle{\\lim_{n \\to \\infty}\\left(\\simplify[std]{1+{a3}/({b3}n)}\\right)^n=} $ [[0]]

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$\\displaystyle{\\lim_{n \\to \\infty}\\frac{\\left(\\simplify[std]{{al^d}n^({a*d})+{be}n^{b}+{c}}\\right)^{1/\\var{d}}} {\\left(\\simplify[std]{{ga^d1}n^({a*d1})+{de}n^{b1}+{c1}}\\right)^{1/\\var{d1}}}=} $ [[0]]

\n

Enter your answer as a fraction or integer, not as a decimal.

", "showCorrectAnswer": true, "marks": 0}], "statement": "

What are the following limits?

", "tags": ["checked2015", "examples of standard limits", "limit", "limits", "limits of sequences", "MAS1601", "sequences", "standard limits", "taking the limit"], "rulesets": {"std": ["all", "fractionNumbers", "!collectNumbers", "!noLeadingMinus"]}, "preamble": {"css": "", "js": ""}, "type": "question", "metadata": {"notes": "\n\t\t

4/07/2012:

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Added tags.

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Improved display of prompt for fourth part.

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Improved display of solution to fourth part.

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Checked calculations.

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No tolerance on answer to 6th part, got to be exact to 4dps. Tolerance variable, tol=0.

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21/07/2012:

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Added description.

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27/7/2012:

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Added tags.

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Question appears to be working correctly.

\n\t\t", "licence": "Creative Commons Attribution 4.0 International", "description": "

Seven standard elementary limits of sequences. 

"}, "variablesTest": {"condition": "", "maxRuns": 100}, "advice": "

All calculations below are rounded to $5$ decimal places.

\n

The notation $a \\approx b$ means that $a$ and $b$ are approximately equal.

\n

a)

\n

Using a calculator for $3$ values of $n$:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
$n$$\\displaystyle{\\frac{1}{n^{1/\\var{r}}}}$
$100$$\\var{v15}$
$5000$$\\var{v110}$
$5000000$$\\var{v150}$
\n

This indicates that $\\displaystyle{\\lim_{n \\to \\infty}\\left(\\frac{1}{n^{1/\\var{r}}}\\right)=0}$

\n

In fact $\\displaystyle{\\lim_{n \\to \\infty}\\left(\\frac{1}{n^r}\\right)=0}$ for any $r \\gt 0$

\n

b)

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
$n$$\\displaystyle{\\var{k1}^{1/n}}$
$100$$\\var{v25}$
$5000$$\\var{v210}$
$5000000$$\\var{v250}$
\n

This indicates that $\\displaystyle{\\lim_{n \\to \\infty}\\var{k1}^{1/n}=1}$.

\n

c)

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
$n$$\\displaystyle{\\var{k}^{1/n}}$
$100$$\\var{v35}$
$5000$$\\var{v310}$
$5000000$$\\var{v350}$
\n

This indicates that $\\displaystyle{\\lim_{n \\to \\infty}\\var{k}^{1/n}=1}$.

\n

From the last two questions it seems that $\\displaystyle{\\lim_{n \\to \\infty} k^{1/n}=1}$ for any $k \\gt 0$ – and this is in fact true.

\n

d)

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
$n$$\\displaystyle{\\frac{\\var{c}n+\\var{d}}{\\var{al}n-\\var{ga}}}$
$100$$\\var{v45}$
$5000$$\\var{v410}$
$5000000$$\\var{v450}$
\n

This indicates that $\\displaystyle \\lim_{n \\to \\infty}\\left(\\simplify[std]{({c}n+{d})/({al}n-{ga})}\\right) = \\simplify[std]{{c}/{al}}$.

\n

In general, $\\displaystyle{ \\lim_{n \\to \\infty}\\left(\\frac{an+b}{cn+d}\\right)= \\frac{a}{c} }$ when $c \\neq 0$.

\n

e)

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
$n$$\\displaystyle{\\left(\\simplify{{c}/{n}}\\right)^n}$
$10$$\\var{v55}$
$29$$\\var{v510}$
$50$$\\var{v550}$
$89$$\\var{v560}$
\n

This indicates that $\\displaystyle{\\lim_{n \\to \\infty}\\left(\\simplify{{c}/{n}}\\right)^n}= 0$.

\n

In general $\\displaystyle{\\lim_{n \\to \\infty} r^n= 0}$ if $|r| \\lt 1$.

\n

f)

\n

We have the limit:

\n

\\[ \\lim_{n\\to\\infty}\\left(1+\\frac{a}{n}\\right) = e^a \\]

\n

The following table confirms that the values are converging to $\\displaystyle{\\simplify[std]{e^({a3}/{b3})={valexp}}}$ (to five decimal places).

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
$n$$\\displaystyle{\\left(\\simplify[std]{1+{a3}/({b3}n)}\\right)^n}$
$10$$\\var{v65}$
$100$$\\var{v610}$
$1000$$\\var{v650}$
$10000$$\\var{v660}$
\n

Hence the answer asked for is $\\var{val}$ to $4$ decimal places.

\n

g)

\n

The answer to this question is based upon neglecting terms in polynomials in $n$ for large $n$.

\n

For example, $n^3+1000000n^2+1000000000 \\approx n^3$ for large $n$ because the $n^3$ term completely dominates the other terms as $n \\to \\infty$.

\n

A more precise way of saying this is:

\n

\\[\\lim_{n\\to\\infty}\\left(\\frac{n^3+1000000n^2+1000000000}{n^3}\\right)=1\\]

\n

So for large $n$,

\n

\\begin{align}
\\frac{\\left(\\simplify[std]{{al^d}n^({a*d})+{be}n^{b}+{c}}\\right)^{1/\\var{d}}} {\\left(\\simplify[std]{{ga^d1}n^({a*d1})+{de}n^{b1}+{c1}}\\right)^{1/\\var{d1}}} &\\approx \\frac{\\left(\\simplify[std]{{al^d}n^({a*d})}\\right)^{1/\\var{d}}} {\\left(\\simplify[std]{{ga^d1}n^({a*d1})}\\right)^{1/\\var{d1}}} \\\\ \\\\
&= \\frac{\\simplify[std]{{al^d}^(1/{d})n^{a}}} {\\simplify[std]{{ga^d1}^(1/{d1})n^{a}}} \\\\ \\\\
&= \\simplify[std]{{al}/{ga}}
\\end{align}

\n

Hence the limit is $\\displaystyle{\\simplify[std]{{al}/{ga}}}$

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