As {choice2[0]}, it appears {choice2[2]} approaches {choice2[1]}, but it isn't clear what this means. This is known as an indeterminate form of type {choice2[1]}. L'Hospital's rule says that if we have such an indeterminate form we can differentiate the numerator and denominator separately and take the limit of the quotient. That is, if $f$ and $g$ are differentiable on an open interval containing $a$, $g'(x)\\ne 0$ on that interval (except possibly at $x=a$), $\\lim_{x\\rightarrow a}f(x)=\\infty$, $\\lim_{x\\rightarrow a}g(x)=\\infty$ and $\\lim_{x\\rightarrow a}\\frac{f'(x)}{g'(x)}$ exists (or equals $\\pm\\infty)$ then:
\n\\[\\lim_{x\\rightarrow a} \\frac{f(x)}{g(x)}=\\lim_{x\\rightarrow a}\\frac{f'(x)}{g'(x)}.\\]
\n(If this is still an indeterminate form we can (of course) repeat the procedure)
\n\nSo by applying L'Hospital's rule we now need to determine what {choice2[3]} approaches as {choice2[0]}. But this is again an indeterminate form so we repeatedly apply L'Hospital's rule until it is not an indeterminate form and we arrive at needing to determine what {choice2[4]} approaches as {choice2[0]}. Now that this is no longer an indeterminate form, it should be clear that the limit is $\\var{choice2[-1]}$.
\n", "tags": [], "extensions": [], "name": "Maria's copy of Limits: L'Hospital's rule: Indeterminate form infinity/infinity", "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["pL", "pa", "pb", "pc", "pd", "pe", "nL", "na", "nb", "nc", "nd", "ne", "log_power", "power_log", "exp_power", "power_exp", "exp_log", "log_exp", "log_root", "root_log", "choice1", "choice2"], "parts": [{"type": "gapfill", "showCorrectAnswer": true, "variableReplacements": [], "gaps": [{"correctAnswerFraction": false, "type": "numberentry", "minValue": "choice2[-1]", "allowFractions": false, "mustBeReducedPC": 0, "scripts": {}, "mustBeReduced": false, "variableReplacementStrategy": "originalfirst", "notationStyles": ["plain", "en", "si-en"], "maxValue": "choice2[-1]", "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "marks": 1, "correctAnswerStyle": "plain"}], "variableReplacementStrategy": "originalfirst", "scripts": {}, "showFeedbackIcon": true, "marks": 0, "prompt": "As {choice2[0]}, {choice2[2]} approaches [[0]]
\nNote: infinity is simply entered by typing infinity
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"}, "pc": {"templateType": "anything", "definition": "pl[2]", "name": "pc", "group": "Ungrouped variables", "description": ""}, "choice2": {"templateType": "anything", "definition": "switch(choice1=0,random(log_power),choice1=1,random(power_log),choice1=2,random(exp_power),choice1=3,random(power_exp),choice1=4,random(exp_log),choice1=5,random(log_exp),choice1=6,random(log_root),choice1=7,random(root_log),'')", "name": "choice2", "group": "Ungrouped variables", "description": ""}, "nc": {"templateType": "anything", "definition": "nl[2]", "name": "nc", "group": "Ungrouped variables", "description": ""}, "pb": {"templateType": "anything", "definition": "pl[1]", "name": "pb", "group": "Ungrouped variables", "description": ""}, "power_log": {"templateType": "anything", "definition": "[\n['\\$x\\\\rightarrow\\\\infty\\$','\\$\\\\frac{\\\\infty}{\\\\infty}\\$','\\$\\\\simplify{({pd}x^{pb}+{nd})/({pc}ln({pa}x+{nb}))}\\$','\\$\\\\simplify{({pd*pb}x^{pb-1})/({pc*pa}/({pa}x+{nb}))=(({pa}x+{nb})*({pd*pb}x^{pb-1}))/({pc*pa})}\\$',infinity], \n['\\$x\\\\rightarrow\\\\infty\\$','\\$\\\\frac{-\\\\infty}{\\\\infty}\\$','\\$\\\\simplify{({-pd}x^{pb}+{nd})/({pc}ln({pa}x+{nb}))}\\$','\\$\\\\simplify{({-pd*pb}x^{pb-1})/({pc*pa}/({pa}x+{nb}))=(({pa}x+{nb})*({-pd*pb}x^{pb-1}))/({pc*pa})}\\$',-infinity], \n\n\n['\\$x\\\\rightarrow-\\\\infty\\$',if(mod(pb,2)=0,'\\$\\\\frac{-\\\\infty}{\\\\infty}\\$','\\$\\\\frac{\\\\infty}{\\\\infty}\\$'), '\\$\\\\simplify{({-pd}x^{pb}+{nd})/({pc}ln({-pa}x+{nb}))}\\$','\\$\\\\simplify{({-pd*pb}x^{pb-1})/({-pc*pa}/({-pa}x+{nb}))=(({-pa}x+{nb})*({-pd*pb}x^{pb-1}))/({-pc*pa})}\\$',if(mod(pb,2)=0,-infinity,infinity)],\n['\\$x\\\\rightarrow-\\\\infty\\$',if(mod(pb,2)=0,'\\$\\\\frac{\\\\infty}{\\\\infty}\\$','\\$\\\\frac{-\\\\infty}{\\\\infty}\\$'), '\\$\\\\simplify{({pd}x^{pb}+{nd})/({pc}ln({-pa}x+{nb}))}\\$','\\$\\\\simplify{({pd*pb}x^{pb-1})/({-pc*pa}/({-pa}x+{nb}))=(({-pa}x+{nb})*({pd*pb}x^{pb-1}))/({-pc*pa})}\\$',if(mod(pb,2)=0,infinity,-infinity)]\n\n\n]\n", "name": "power_log", "group": "Ungrouped variables", "description": "using a list to keep track of important things
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"}}, "metadata": {"licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International", "description": "Just what the title says, I guess.
"}, "functions": {}, "variable_groups": [], "preamble": {"css": "", "js": ""}, "rulesets": {}, "statement": "This question is about limits of indeterminate forms.
", "type": "question", "contributors": [{"name": "Ben Brawn", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/605/"}, {"name": "Maria Aneiros", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3388/"}]}]}], "contributors": [{"name": "Ben Brawn", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/605/"}, {"name": "Maria Aneiros", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3388/"}]}