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$\\displaystyle{\\lim_{n \\to \\infty}\\left(\\frac{1}{n^{1/\\var{r}}}\\right)=} $ [[0]]
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", "showCorrectAnswer": true, "marks": 0}, {"scripts": {}, "gaps": [{"showCorrectAnswer": true, "allowFractions": false, "scripts": {}, "type": "numberentry", "maxValue": "1", "minValue": "1", "correctAnswerFraction": false, "marks": 1, "showPrecisionHint": false}], "type": "gapfill", "prompt": "$\\displaystyle{\\lim_{n \\to \\infty}\\left(\\var{k}^{1/n}\\right)=} $ [[0]]
", "showCorrectAnswer": true, "marks": 0}, {"scripts": {}, "gaps": [{"showCorrectAnswer": true, "allowFractions": true, "scripts": {}, "type": "numberentry", "maxValue": "c/al", "minValue": "c/al", "correctAnswerFraction": true, "marks": 1, "showPrecisionHint": false}], "type": "gapfill", "prompt": "$\\displaystyle \\lim_{n \\to \\infty}\\left(\\simplify[std]{({c}n+{d})/({al}n-{ga})}\\right) = $ [[0]]
\nEnter your answer as a fraction or integer, not as a decimal.
", "showCorrectAnswer": true, "marks": 0}, {"scripts": {}, "gaps": [{"showCorrectAnswer": true, "allowFractions": false, "scripts": {}, "type": "numberentry", "maxValue": "0", "minValue": "0", "correctAnswerFraction": false, "marks": 1, "showPrecisionHint": false}], "type": "gapfill", "prompt": "$\\displaystyle{\\lim_{n \\to \\infty}\\left(\\simplify[std]{{c}/{n}}\\right)^n=} $ [[0]]
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", "showCorrectAnswer": true, "marks": 0}, {"scripts": {}, "gaps": [{"showCorrectAnswer": true, "allowFractions": true, "scripts": {}, "type": "numberentry", "maxValue": "al/ga", "minValue": "al/ga", "correctAnswerFraction": true, "marks": 1, "showPrecisionHint": false}], "type": "gapfill", "prompt": "$\\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]]
\nEnter 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\t4/07/2012:
\n\t\tAdded tags.
\n\t\tImproved display of prompt for fourth part.
\n\t\tImproved display of solution to fourth part.
\n\t\tChecked calculations.
\n\t\tNo tolerance on answer to 6th part, got to be exact to 4dps. Tolerance variable, tol=0.
\n\t\t21/07/2012:
\n\t\tAdded description.
\n\t\t27/7/2012:
\n\t\tAdded tags.
\n\t\tQuestion 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.
\nThe notation $a \\approx b$ means that $a$ and $b$ are approximately equal.
\nUsing a calculator for $3$ values of $n$:
\n$n$ | $\\displaystyle{\\frac{1}{n^{1/\\var{r}}}}$ |
---|---|
$100$ | \n$\\var{v15}$ | \n
$5000$ | \n$\\var{v110}$ | \n
$5000000$ | \n$\\var{v150}$ | \n
This indicates that $\\displaystyle{\\lim_{n \\to \\infty}\\left(\\frac{1}{n^{1/\\var{r}}}\\right)=0}$
\nIn fact $\\displaystyle{\\lim_{n \\to \\infty}\\left(\\frac{1}{n^r}\\right)=0}$ for any $r \\gt 0$
\n$n$ | $\\displaystyle{\\var{k1}^{1/n}}$ |
---|---|
$100$ | \n$\\var{v25}$ | \n
$5000$ | \n$\\var{v210}$ | \n
$5000000$ | \n$\\var{v250}$ | \n
This indicates that $\\displaystyle{\\lim_{n \\to \\infty}\\var{k1}^{1/n}=1}$.
\n$n$ | $\\displaystyle{\\var{k}^{1/n}}$ |
---|---|
$100$ | \n$\\var{v35}$ | \n
$5000$ | \n$\\var{v310}$ | \n
$5000000$ | \n$\\var{v350}$ | \n
This indicates that $\\displaystyle{\\lim_{n \\to \\infty}\\var{k}^{1/n}=1}$.
\nFrom 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$n$ | $\\displaystyle{\\frac{\\var{c}n+\\var{d}}{\\var{al}n-\\var{ga}}}$ |
---|---|
$100$ | \n$\\var{v45}$ | \n
$5000$ | \n$\\var{v410}$ | \n
$5000000$ | \n$\\var{v450}$ | \n
This indicates that $\\displaystyle \\lim_{n \\to \\infty}\\left(\\simplify[std]{({c}n+{d})/({al}n-{ga})}\\right) = \\simplify[std]{{c}/{al}}$.
\nIn general, $\\displaystyle{ \\lim_{n \\to \\infty}\\left(\\frac{an+b}{cn+d}\\right)= \\frac{a}{c} }$ when $c \\neq 0$.
\n$n$ | $\\displaystyle{\\left(\\simplify{{c}/{n}}\\right)^n}$ |
---|---|
$10$ | \n$\\var{v55}$ | \n
$29$ | \n$\\var{v510}$ | \n
$50$ | \n$\\var{v550}$ | \n
$89$ | \n$\\var{v560}$ | \n
This indicates that $\\displaystyle{\\lim_{n \\to \\infty}\\left(\\simplify{{c}/{n}}\\right)^n}= 0$.
\nIn general $\\displaystyle{\\lim_{n \\to \\infty} r^n= 0}$ if $|r| \\lt 1$.
\nWe have the limit:
\n\\[ \\lim_{n\\to\\infty}\\left(1+\\frac{a}{n}\\right) = e^a \\]
\nThe following table confirms that the values are converging to $\\displaystyle{\\simplify[std]{e^({a3}/{b3})={valexp}}}$ (to five decimal places).
\n$n$ | $\\displaystyle{\\left(\\simplify[std]{1+{a3}/({b3}n)}\\right)^n}$ |
---|---|
$10$ | \n$\\var{v65}$ | \n
$100$ | \n$\\var{v610}$ | \n
$1000$ | \n$\\var{v650}$ | \n
$10000$ | \n$\\var{v660}$ | \n
Hence the answer asked for is $\\var{val}$ to $4$ decimal places.
\nThe answer to this question is based upon neglecting terms in polynomials in $n$ for large $n$.
\nFor example, $n^3+1000000n^2+1000000000 \\approx n^3$ for large $n$ because the $n^3$ term completely dominates the other terms as $n \\to \\infty$.
\nA more precise way of saying this is:
\n\\[\\lim_{n\\to\\infty}\\left(\\frac{n^3+1000000n^2+1000000000}{n^3}\\right)=1\\]
\nSo 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}
Hence the limit is $\\displaystyle{\\simplify[std]{{al}/{ga}}}$
", "contributors": [{"name": "Newcastle University Mathematics and Statistics", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/697/"}]}]}], "contributors": [{"name": "Newcastle University Mathematics and Statistics", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/697/"}]}