Just what the title says, I guess. I couldn't find a 0^0 that didn't converge to 1 except things like x^(1/ln(x)) as x->0, but they just need the e^ln() transformation, not L'hopital's rule!
"}, "variables": {"one_inf": {"name": "one_inf", "group": "Ungrouped variables", "description": "", "definition": "[\n['\\$x\\\\rightarrow\\\\infty\\$','\\$1^{\\\\infty}\\$','\\$\\\\simplify{(1+{na}/x)^({pa}x+{nb})}\\$','\\$\\\\infty\\\\cdot 0\\$','\\$\\\\simplify{(1+{na}/x)^({pa}x+{nb})=e^(ln((1+{na}/x)^({pa}x+{nb})))=e^(({pa}x+{nb})*ln(1+{na}/x))}\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{(1+{na}/x)^({pa}x+{nb})=e^(({pa}x+{nb})*ln(1+{na}/x))=e^(ln(1+{na}/x)/({pa}x+{nb})^(-1))}\\$','\\$\\\\simplify{e^(((-{na}x^-2)/(ln(1+{na}x^(-1))))/(-{pa}({pa}x+{nb})^(-2)))=e^(({na*pa^2}+{2*na*pa*nb}/x+{nb^2}/x^2)/({pa}+{pa*na}/x))}\\$',e^(na*pa),'\\$\\\\simplify[!otherNumbers]{e^({na*pa^2}/{pa})=e^({na*pa})}\\$'],\n['\\$x\\\\rightarrow\\\\infty\\$','\\$1^{\\\\infty}\\$','\\$\\\\simplify{((x+{na})/(x+{nb}))^({nc}x)}\\$','\\$\\\\infty\\\\cdot 0\\$','\\$\\\\simplify{((x+{na})/(x+{nb}))^({nc}x)=e^(ln(((x+{na})/(x+{nb}))^({nc}x)))=e^({nc}x*ln((x+{na})/(x+{nb})))}\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{((x+{na})/(x+{nb}))^({nc}x)=e^({nc}x*ln((x+{na})/(x+{nb})))=e^(({nc}ln((x+{na})/(x+{nb})))/(x^(-1)))}\\$','\\$\\\\simplify{e^(({nc}({nb}-{na})/((x+{na})(x+{nb})))/(-x^(-2)))=e^(({nc*(na-nb)}x^2)/(x^2+{na+nb}x+{na*nb}))}\\$',e^(nc*(na-nb)),'\\$\\\\simplify[!otherNumbers]{e^{nc*(na-nb)}}\\$'],\n['\\$x\\\\rightarrow 0^+\\$','\\$1^{\\\\infty}\\$','\\$\\\\simplify{(1+{na}x)^(({pa}+{nb}x)/x)}\\$','\\$\\\\infty\\\\cdot 0\\$','\\$\\\\simplify{(1+{na}x)^(({pa}+{nb}x)/x)=e^(ln((1+{na}x)^(({pa}+{nb}x)/x)))=e^((({pa}+{nb}x)/x*ln(1+{na}x)))}\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{(1+{na}x)^(({pa}+{nb}x)/x)=e^((({pa}+{nb}x)/x*ln(1+{na}x)))=e^((({pa}+{nb}x)*ln(1+{na}x))/x)}\\$','\\$\\\\simplify{e^({nb}ln(1+{na}x)+{na}({pa}+{nb}x)/(1+{na}x))}\\$',e^(na*pa),'\\$\\\\simplify[!otherNumbers]{e^({na*pa})}\\$'],\n['\\$x\\\\rightarrow 0^-\\$','\\$1^{-\\\\infty}\\$','\\$\\\\simplify{(1+{na}x)^(({pa}+{nb}x)/x)}\\$','\\$-\\\\infty\\\\cdot 0\\$','\\$\\\\simplify{(1+{na}x)^(({pa}+{nb}x)/x)=e^(ln((1+{na}x)^(({pa}+{nb}x)/x)))=e^((({pa}+{nb}x)/x*ln(1+{na}x)))}\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{(1+{na}x)^(({pa}+{nb}x)/x)=e^((({pa}+{nb}x)/x*ln(1+{na}x)))=e^((({pa}+{nb}x)*ln(1+{na}x))/x)}\\$','\\$\\\\simplify{e^({nb}ln(1+{na}x)+{na}({pa}+{nb}x)/(1+{na}x))}\\$',e^(na*pa),'\\$\\\\simplify[!otherNumbers]{e^({na*pa})}\\$'],\n['\\$x\\\\rightarrow \\\\var{nb}^+\\$',if(na>0,'\\$1^{\\\\infty}\\$','\\$1^{-\\\\infty}\\$'),'\\$\\\\simplify{(x-{nb-1})^({na}/(x-{nb}))}\\$',if(na>0,'\\$\\\\infty\\\\cdot 0\\$','\\$-\\\\infty\\\\cdot 0\\$'),'\\$\\\\simplify{(x-{nb-1})^({na}/(x-{nb}))=e^(ln((x-{nb-1})^({na}/(x-{nb}))))=e^(({na}/(x-{nb}))*ln(x-{nb-1}))}\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{(x-{nb-1})^({na}/(x-{nb}))=e^(ln((x-{nb-1})^({na}/(x-{nb}))))=e^(({na}/(x-{nb}))*ln(x-{nb-1}))=e^({na}*ln(x-{nb-1})/(x-{nb}))}\\$','\\$\\\\simplify{e^({na}/(x-{nb-1}))}\\$',e^(na),'\\$\\\\simplify[!otherNumbers]{e^{na}}\\$'],\n['\\$x\\\\rightarrow \\\\var{nb}^-\\$',if(na<0,'\\$1^{\\\\infty}\\$','\\$1^{-\\\\infty}\\$'),'\\$\\\\simplify{(x-{nb-1})^({na}/(x-{nb}))}\\$',if(na<0,'\\$\\\\infty\\\\cdot 0\\$','\\$-\\\\infty\\\\cdot 0\\$'),'\\$\\\\simplify{(x-{nb-1})^({na}/(x-{nb}))=e^(ln((x-{nb-1})^({na}/(x-{nb}))))=e^(({na}/(x-{nb}))*ln(x-{nb-1}))}\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{(x-{nb-1})^({na}/(x-{nb}))=e^(ln((x-{nb-1})^({na}/(x-{nb}))))=e^(({na}/(x-{nb}))*ln(x-{nb-1}))=e^({na}*ln(x-{nb-1})/(x-{nb}))}\\$','\\$\\\\simplify{e^({na}/(x-{nb-1}))}\\$',e^(na),'\\$\\\\simplify[!otherNumbers]{e^{na}}\\$']\n //still working on the above \n]", "templateType": "anything"}, "zero_zero": {"name": "zero_zero", "group": "Ungrouped variables", "description": "", "definition": "[\n\n['\\$x\\\\rightarrow 0\\$','\\$0^0\\$','\\$\\\\simplify{x^x}\\$','\\$0\\\\cdot -\\\\infty\\$','\\$\\\\simplify{x^x=e^(ln(x^x))=e^(x*ln(x))}\\$','\\$\\\\frac{-\\\\infty}{\\\\infty}\\$','\\$\\\\simplify{x^x=e^(ln(x^x))=e^(x*ln(x))=e^(ln(x)/(1/x))}\\$','\\$\\\\simplify{e^((1/x)/(-1/x^2))=e^(-x)}\\$',1,'\\$ 1 \\$'],\n['\\$x\\\\rightarrow \\\\var{na}\\$','\\$0^0\\$','\\$\\\\simplify{(x-{na})^(x-{na})}\\$','\\$0\\\\cdot -\\\\infty\\$','\\$\\\\simplify{(x-{na})^(x-{na})=e^(ln((x-{na})^(x-{na})))=e^((x-{na})*ln(x-{na}))}\\$','\\$\\\\frac{-\\\\infty}{\\\\infty}\\$','\\$\\\\simplify{(x-{na})^(x-{na})=e^(ln((x-{na})^(x-{na})))=e^((x-{na})*ln(x-{na}))=e^(ln(x-{na})/(1/(x-{na})))}\\$','\\$\\\\simplify{e^((1/(x-{na}))/(-1/(x-{na})^2))=e^({na}-x)}\\$',1,'\\$ 1 \\$'],\n\n['\\$x\\\\rightarrow 0\\$','\\$0^0\\$','\\$\\\\simplify{sin(x)^sin(x)}\\$','\\$0\\\\cdot -\\\\infty\\$','\\$\\\\simplify{sin(x)^sin(x)=e^(ln(sin(x)^sin(x)))=e^(sin(x)*ln(sin(x)))}\\$','\\$\\\\frac{-\\\\infty}{\\\\infty}\\$','\\$\\\\simplify{sin(x)^sin(x)=e^(ln(sin(x)^sin(x)))=e^(sin(x)*ln(sin(x)))=e^(ln(sin(x))/(1/sin(x)))}\\$','\\$\\\\simplify{e^((cos(x)/sin(x))/(-cos(x)/sin(x)^2))=e^(-sin(x))}\\$',1,'\\$ 1 \\$'],\n['\\$x\\\\rightarrow \\\\var{na}\\$','\\$0^0\\$','\\$\\\\simplify{sin(x-{na})^sin(x-{na})}\\$','\\$0\\\\cdot -\\\\infty\\$','\\$\\\\simplify{sin(x-{na})^sin(x-{na})=e^(ln(sin(x-{na})^sin(x-{na})))=e^(sin(x-{na})*ln(sin(x-{na})))}\\$','\\$\\\\frac{-\\\\infty}{\\\\infty}\\$','\\$\\\\simplify{sin(x-{na})^sin(x-{na})=e^(ln(sin(x-{na})^sin(x-{na})))=e^(sin(x-{na})*ln(sin(x-{na})))=e^(ln(sin(x-{na}))/(1/sin(x-{na})))}\\$','\\$\\\\simplify{e^((cos(x-{na})/sin(x-{na}))/(-cos(x-{na})/sin(x-{na})^2))=e^(-sin(x-{na}))}\\$',1,'\\$ 1 \\$'], \n \n['\\$x\\\\rightarrow \\\\frac{\\\\pi}{2}\\$','\\$0^0\\$','\\$\\\\simplify{cos(x)^cos(x)}\\$','\\$0\\\\cdot -\\\\infty\\$','\\$\\\\simplify{cos(x)^cos(x)=e^(ln(cos(x)^cos(x)))=e^(cos(x)*ln(cos(x)))}\\$','\\$\\\\frac{-\\\\infty}{\\\\infty}\\$','\\$\\\\simplify{cos(x)^cos(x)=e^(ln(cos(x)^cos(x)))=e^(cos(x)*ln(cos(x)))=e^(ln(cos(x))/(1/cos(x)))}\\$','\\$\\\\simplify{e^((-sin(x)/cos(x))/(--sin(x)/cos(x)^2))=e^(-cos(x))}\\$',1,'\\$ 1 \\$'] , \n \n \n['\\$x\\\\rightarrow 0\\$','\\$0^0\\$','\\$\\\\simplify{sin(x)^tan(x)}\\$','\\$0\\\\cdot -\\\\infty\\$','\\$\\\\simplify{sin(x)^tan(x)=e^(ln(sin(x)^tan(x)))=e^(tan(x)*ln(sin(x)))}\\$','\\$\\\\frac{-\\\\infty}{\\\\infty}\\$','\\$\\\\simplify{sin(x)^tan(x)=e^(ln(sin(x)^tan(x)))=e^(tan(x)*ln(sin(x)))=e^(ln(sin(x))/(cot(x)))}\\$','\\$\\\\simplify{e^((cos(x)/sin(x))/(-cosec(x)^2))=e^(-sin(x)*cos(x))}\\$',1,'\\$ 1 \\$'],\n['\\$x\\\\rightarrow \\\\var{na}\\$','\\$0^0\\$','\\$\\\\simplify{sin(x-{na})^tan(x-{na})}\\$','\\$0\\\\cdot -\\\\infty\\$','\\$\\\\simplify{sin(x-{na})^tan(x-{na})=e^(ln(sin(x-{na})^tan(x-{na})))=e^(tan(x-{na})*ln(sin(x-{na})))}\\$','\\$\\\\frac{-\\\\infty}{\\\\infty}\\$','\\$\\\\simplify{sin(x-{na})^tan(x-{na})=e^(ln(sin(x-{na})^tan(x-{na})))=e^(tan(x-{na})*ln(sin(x-{na})))=e^(ln(sin(x-{na}))/(cot(x-{na})))}\\$','\\$\\\\simplify{e^((cos(x-{na})/sin(x-{na}))/(-cosec(x-{na})^2))=e^(-sin(x-{na})*cos(x-{na}))}\\$',1,'\\$ 1 \\$'],\n \n['\\$x\\\\rightarrow 0\\$','\\$0^0\\$','\\$\\\\simplify{tan(x)^tan(x)}\\$','\\$0\\\\cdot -\\\\infty\\$','\\$\\\\simplify{tan(x)^tan(x)=e^(ln(tan(x)^tan(x)))=e^(tan(x)*ln(tan(x)))}\\$','\\$\\\\frac{-\\\\infty}{\\\\infty}\\$','\\$\\\\simplify{tan(x)^tan(x)=e^(ln(tan(x)^tan(x)))=e^(tan(x)*ln(tan(x)))=e^(ln(tan(x))/(cot(x)))}\\$','\\$\\\\simplify{e^((sec(x)^2/tan(x))/(-cosec(x)^2))=e^(-tan(x))}\\$',1,'\\$1\\$'],\n['\\$x\\\\rightarrow \\\\var{na}\\$','\\$0^0\\$','\\$\\\\simplify{(tan(x-{na}))^tan(x-{na})}\\$','\\$0\\\\cdot -\\\\infty\\$','\\$\\\\simplify{tan(x-{na})^tan(x-{na})=e^(ln(tan(x-{na})^tan(x-{na})))=e^(tan(x-{na})*ln(tan(x-{na})))}\\$','\\$\\\\frac{-\\\\infty}{\\\\infty}\\$','\\$\\\\simplify{tan(x-{na})^tan(x-{na})=e^(ln(tan(x-{na})^tan(x-{na})))=e^(tan(x-{na})*ln(tan(x-{na})))=e^(ln(tan(x-{na}))/(cot(x-{na})))}\\$','\\$\\\\simplify{e^((sec(x-{na})^2/tan(x-{na}))/(-cosec(x-{na})^2))=e^(-tan(x-{na}))}\\$',1,'\\$1\\$']\n \n, \n \n['\\$x\\\\rightarrow 0\\$','\\$0^0\\$','\\$\\\\simplify{x^({na}x)}\\$','\\$0\\\\cdot -\\\\infty\\$','\\$\\\\simplify{x^({na}x)=e^(ln(x^({na}x)))=e^(({na}x)*ln(x))}\\$','\\$\\\\frac{-\\\\infty}{\\\\infty}\\$','\\$\\\\simplify{x^({na}x)=e^(ln(x^({na}x)))=e^(({na}x)*ln(x))=e^(ln(x)/(1/({na}x)))}\\$','\\$\\\\simplify{e^((1/x)/(-{na}/({na}x)^2))=e^(-{na}x)}\\$',1,'\\$ 1 \\$'],\n\n['\\$x\\\\rightarrow 0\\$','\\$0^0\\$','\\$\\\\simplify{sin({nb}x)^sin({nb}x)}\\$','\\$0\\\\cdot -\\\\infty\\$','\\$\\\\simplify{sin({nb}x)^sin({nb}x)=e^(ln(sin({nb}x)^sin({nb}x)))=e^(sin({nb}x)*ln(sin({nb}x)))}\\$','\\$\\\\frac{-\\\\infty}{\\\\infty}\\$','\\$\\\\simplify{sin({nb}x)^sin({nb}x)=e^(ln(sin({nb}x)^sin({nb}x)))=e^(sin({nb}x)*ln(sin({nb}x)))=e^(ln(sin({nb}x))/(1/sin({nb}x)))}\\$','\\$\\\\simplify{e^(({nb}cos({nb}x)/sin({nb}x))/(-{nb}cos({nb}x)/sin({nb}x)^2))=e^(-sin(x))}\\$',1,'\\$ 1 \\$'],\n \n['\\$x\\\\rightarrow \\\\frac{\\\\pi}{\\\\var{2*pa}}\\$','\\$0^0\\$','\\$\\\\simplify{cos({pa}x)^cos({pa}x)}\\$','\\$0\\\\cdot -\\\\infty\\$','\\$\\\\simplify{cos({pa}x)^cos({pa}x)=e^(ln(cos({pa}x)^cos({pa}x)))=e^(cos({pa}x)*ln(cos({pa}x)))}\\$','\\$\\\\frac{-\\\\infty}{\\\\infty}\\$','\\$\\\\simplify{cos({pa}x)^cos({pa}x)=e^(ln(cos({pa}x)^cos({pa}x)))=e^(cos({pa}x)*ln(cos({pa}x)))=e^(ln(cos({pa}x))/(1/cos({pa}x)))}\\$','\\$\\\\simplify{e^((-sin({pa}x)/cos({pa}x))/(--sin({pa}x)/cos({pa}x)^2))=e^(-cos({pa}x))}\\$',1,'\\$ 1 \\$'] , \n\n \n['\\$x\\\\rightarrow 0\\$','\\$0^0\\$','\\$\\\\simplify{sin({na}x)^tan({na}x)}\\$','\\$0\\\\cdot -\\\\infty\\$','\\$\\\\simplify{sin({na}x)^tan({na}x)=e^(ln(sin({na}x)^tan({na}x)))=e^(tan({na}x)*ln(sin({na}x)))}\\$','\\$\\\\frac{-\\\\infty}{\\\\infty}\\$','\\$\\\\simplify{sin({na}x)^tan({na}x)=e^(ln(sin({na}x)^tan({na}x)))=e^(tan({na}x)*ln(sin({na}x)))=e^(ln(sin({na}x))/(cot({na}x)))}\\$','\\$\\\\simplify{e^(({na}cos({na}x)/sin({na}x))/(-{na}cosec({na}x)^2))=e^(-sin({na}x)*cos({na}x))}\\$',1,'\\$ 1 \\$'],\n\n['\\$x\\\\rightarrow 0\\$','\\$0^0\\$','\\$\\\\simplify{tan({na}x)^tan({na}x)}\\$','\\$0\\\\cdot -\\\\infty\\$','\\$\\\\simplify{tan({na}x)^tan({na}x)=e^(ln(tan({na}x)^tan({na}x)))=e^(tan({na}x)*ln(tan({na}x)))}\\$','\\$\\\\frac{-\\\\infty}{\\\\infty}\\$','\\$\\\\simplify{tan({na}x)^tan({na}x)=e^(ln(tan({na}x)^tan({na}x)))=e^(tan({na}x)*ln(tan({na}x)))=e^(ln(tan({na}x))/(cot({na}x)))}\\$','\\$\\\\simplify{e^(({na}sec({na}x)^2/tan({na}x))/(-{na}cosec({na}x)^2))=e^(-tan({na}x))}\\$',1,'\\$1\\$']\n]", "templateType": "anything"}, "inf_zero": {"name": "inf_zero", "group": "Ungrouped variables", "description": "using a list to keep track of important things
\nwhat x approaches, indeterminate form type, what y is, the first application of L'Hospital's, (if needed) the final application of L'Hopital's, what y approaches.
", "definition": "[\n\n['\\$x\\\\rightarrow\\\\infty\\$','\\${\\\\infty}^{0}\\$','\\$\\\\simplify{(e^x+{na})^({nb}/x)}\\$','\\$0\\\\cdot \\\\infty\\$','\\$\\\\simplify{(e^x+{na})^({nb}/x)=e^(({nb}/x)*ln(e^x+{na}))}\\$','\\$\\\\frac{\\\\var{nb*infinity}}{\\\\infty}\\$','\\$\\\\simplify{(e^x+{na})^({nb}/x)=e^(({nb}/x)*ln(e^x+{na}))=e^({nb}ln(e^x+{na})/x)}\\$','\\$\\\\simplify{e^({nb}e^x/(e^x+{na}))}\\$','\\$\\\\simplify[!otherNumbers]{e^({nb}e^x/(e^x))=e^({nb})}\\$',e^nb,'\\$e^{\\\\var{nb}}\\$'], \n\n['\\$x\\\\rightarrow 0^+\\$','\\${\\\\infty}^{0}\\$','\\$\\\\simplify{(e^(x^(-1))+{na})^({nb}*x)}\\$','\\$0\\\\cdot \\\\infty\\$','\\$\\\\simplify{(e^(x^(-1))+{na})^({nb}*x)=e^(({nb}*x)*ln(e^(x^(-1))+{na}))}\\$','\\$\\\\frac{\\\\var{nb*infinity}}{\\\\infty}\\$','\\$\\\\simplify{(e^(x^(-1))+{na})^({nb}*x)=e^(({nb}*x)*ln(e^(x^(-1))+{na}))=e^({nb}ln(e^(x^(-1))+{na})/x^(-1))}\\$','\\$\\\\simplify{e^(({nb}(-x^(-2)e^(x^(-1)))/(e^(x^(-1))+{na}))/(-x^(-2)))=e^({nb}e^(x^(-1))/(e^(x^(-1))+{na}))}\\$','\\$\\\\simplify[!otherNumbers]{e^({nb}(-x^(-2)e^(x^(-1)))/(-x^(-2)e^(x^(-1))))=e^({nb})}\\$',e^nb,'\\$e^{\\\\var{nb}}\\$'], \n\n['\\$x\\\\rightarrow \\\\infty\\$','\\${\\\\infty}^{0}\\$','\\$\\\\simplify{(e^x+{na})^(ln({pa})/x)}\\$','\\$0\\\\cdot \\\\infty\\$','\\$\\\\simplify{(e^x+{na})^(ln({pa})/x)=e^((ln({pa})/x)*ln(e^x+{na}))}\\$','\\$\\\\frac{\\\\infty}{\\\\infty}\\$','\\$\\\\simplify{(e^x+{na})^(ln({pa})/x)=e^((ln({pa})/x)*ln(e^x+{na}))=e^(ln({pa})ln(e^x+{na})/x)}\\$','\\$\\\\simplify{e^(ln({pa})e^x/(e^x+{na}))}\\$','\\$\\\\simplify{e^(ln({pa})e^x/(e^x))=e^(ln({pa}))={pa}}\\$',pa,'\\$\\\\var{pa}\\$'], \n\n['\\$x\\\\rightarrow 0^+\\$','\\${\\\\infty}^{0}\\$','\\$\\\\simplify{(e^(x^(-1))+{na})^(ln({pa})x)}\\$','\\$0\\\\cdot \\\\infty\\$','\\$\\\\simplify{(e^(x^(-1))+{na})^(ln({pa})x)=e^((ln({pa})x)*ln(e^(x^(-1))+{na}))}\\$','\\$\\\\frac{\\\\var{nb*infinity}}{\\\\infty}\\$','\\$\\\\simplify{(e^(x^(-1))+{na})^(ln({pa})x)=e^((ln({pa})x)*ln(e^(x^(-1))+{na}))=e^(ln({pa})ln(e^(x^(-1))+{na})/x^(-1))}\\$','\\$\\\\simplify{e^((ln({pa})(-x^(-2)e^(x^(-1)))/(e^(x^(-1))+{na}))/(-x^(-2)))=e^(ln({pa})e^(x^(-1))/(e^(x^(-1))+{na}))}\\$','\\$\\\\simplify{e^(ln({pa})(-x^(-2)e^(x^(-1)))/(-x^(-2)e^(x^(-1))))=e^(ln({pa}))={pa}}\\$',pa,'\\$\\\\var{pa}\\$'], \n\n\n['\\$x\\\\rightarrow 0^+\\$','\\${\\\\infty}^{0}\\$','\\$\\\\simplify{(sinh({pa}/x))^({nb}*x)}\\$','\\$0\\\\cdot \\\\infty\\$','\\$\\\\simplify{(sinh({pa}/x))^({nb}*x)=e^(({nb}*x)*ln(sinh({pa}/x)))}\\$','\\$\\\\frac{\\\\var{nb*infinity}}{\\\\infty}\\$','\\$\\\\simplify{(sinh({pa}/x))^({nb}*x)=e^(({nb}*x)*ln(sinh({pa}/x)))=e^({nb}ln(sinh({pa}/x))/x^(-1))}\\$','\\$\\\\simplify{e^(((-{pa*nb}x^(-2)cosh({pa}/x))/(sinh({pa}/x)))/(-x^(-2)))=e^({pa*nb}coth({pa}/x))}\\$',e^(pa*nb),'\\$e^{\\\\var{pa*nb}}\\$'], \n\n ['\\$x\\\\rightarrow 0^-\\$','\\${\\\\infty}^{0}\\$','\\$\\\\simplify{(sinh({-pa}/x))^({nb}*x)}\\$','\\$0\\\\cdot \\\\infty\\$','\\$\\\\simplify{(sinh({-pa}/x))^({nb}*x)=e^(({nb}*x)*ln(sinh({-pa}/x)))}\\$','\\$\\\\frac{\\\\var{-nb*infinity}}{\\\\infty}\\$','\\$\\\\simplify{(sinh({-pa}/x))^({nb}*x)=e^(({nb}*x)*ln(sinh({-pa}/x)))=e^({nb}ln(sinh({-pa}/x))/x^(-1))}\\$','\\$\\\\simplify{e^((({pa*nb}x^(-2)cosh({-pa}/x))/(sinh({-pa}/x)))/(-x^(-2)))=e^({-pa*nb}coth({-pa}/x))}\\$',e^(-pa*nb),'\\$e^{\\\\var{-pa*nb}}\\$'], \n \n \n['\\$x\\\\rightarrow 0^+\\$','\\${\\\\infty}^{0}\\$','\\$\\\\simplify{(cosh({pa}/x))^({nb}*x)}\\$','\\$0\\\\cdot \\\\infty\\$','\\$\\\\simplify{(cosh({pa}/x))^({nb}*x)=e^(({nb}*x)*ln(cosh({pa}/x)))}\\$','\\$\\\\frac{\\\\var{nb*infinity}}{\\\\infty}\\$','\\$\\\\simplify{(cosh({pa}/x))^({nb}*x)=e^(({nb}*x)*ln(cosh({pa}/x)))=e^({nb}ln(cosh({pa}/x))/x^(-1))}\\$','\\$\\\\simplify{e^(((-{pa*nb}x^(-2)sinh({pa}/x))/(cosh({pa}/x)))/(-x^(-2)))=e^({pa*nb}tanh({pa}/x))}\\$',e^(pa*nb),'\\$e^{\\\\var{pa*nb}}\\$'], \n\n['\\$x\\\\rightarrow 0^-\\$','\\${\\\\infty}^{0}\\$','\\$\\\\simplify{(cosh({pa}/x))^({nb}*x)}\\$','\\$0\\\\cdot \\\\infty\\$','\\$\\\\simplify{(cosh({pa}/x))^({nb}*x)=e^(({nb}*x)*ln(cosh({pa}/x)))}\\$','\\$\\\\frac{\\\\var{nb*infinity}}{-\\\\infty}\\$','\\$\\\\simplify{(cosh({pa}/x))^({nb}*x)=e^(({nb}*x)*ln(cosh({pa}/x)))=e^({nb}ln(cosh({pa}/x))/x^(-1))}\\$','\\$\\\\simplify{e^(((-{pa*nb}x^(-2)sinh({pa}/x))/(cosh({pa}/x)))/(-x^(-2)))=e^({pa*nb}tanh({pa}/x))}\\$',e^(-pa*nb),'\\$e^{\\\\var{-pa*nb}}\\$'], \n\n\n['\\$x\\\\rightarrow \\\\infty\\$','\\${\\\\infty}^{0}\\$','\\$\\\\simplify{(sinh({pa}*x))^({nb}/x)}\\$','\\$0\\\\cdot \\\\infty\\$','\\$\\\\simplify{(sinh({pa}*x))^({nb}/x)=e^(({nb}/x)*ln(sinh({pa}*x)))}\\$','\\$\\\\frac{\\\\var{nb*infinity}}{\\\\infty}\\$','\\$\\\\simplify{(sinh({pa}*x))^({nb}/x)=e^(({nb}/x)*ln(sinh({pa}*x)))=e^({nb}ln(sinh({pa}*x))/x)}\\$','\\$\\\\simplify{e^(({pa*nb}cosh({pa}*x))/(sinh({pa}*x)))=e^({pa*nb}coth({pa}/x))}\\$',e^(pa*nb),'\\$e^{\\\\var{pa*nb}}\\$'], \n\n['\\$x\\\\rightarrow \\\\infty\\$','\\${\\\\infty}^{0}\\$','\\$\\\\simplify{(cosh({pa}*x))^({nb}/x)}\\$','\\$0\\\\cdot \\\\infty\\$','\\$\\\\simplify{(cosh({pa}*x))^({nb}/x)=e^(({nb}/x)*ln(cosh({pa}*x)))}\\$','\\$\\\\frac{\\\\var{nb*infinity}}{\\\\infty}\\$','\\$\\\\simplify{(cosh({pa}*x))^({nb}/x)=e^(({nb}/x)*ln(cosh({pa}*x)))=e^({nb}ln(cosh({pa}*x))/x)}\\$','\\$\\\\simplify{e^(({pa*nb}sinh({pa}*x))/(cosh({pa}*x)))=e^({pa*nb}tanh({pa}/x))}\\$',e^(pa*nb),'\\$e^{\\\\var{pa*nb}}\\$'], \n \n['\\$x\\\\rightarrow -\\\\infty\\$','\\${\\\\infty}^{0}\\$','\\$\\\\simplify{(cosh({pa}*x))^({nb}/x)}\\$','\\$0\\\\cdot \\\\infty\\$','\\$\\\\simplify{(cosh({pa}*x))^({nb}/x)=e^(({nb}/x)*ln(cosh({pa}*x)))}\\$','\\$\\\\frac{\\\\var{nb*infinity}}{-\\\\infty}\\$','\\$\\\\simplify{(cosh({pa}*x))^({nb}/x)=e^(({nb}/x)*ln(cosh({pa}*x)))=e^({nb}ln(cosh({pa}*x))/x)}\\$','\\$\\\\simplify{e^(({pa*nb}sinh({pa}*x))/(cosh({pa}*x)))=e^({pa*nb}tanh({pa}/x))}\\$',e^(-pa*nb),'\\$e^{\\\\var{-pa*nb}}\\$'], \n\n['\\$x\\\\rightarrow -\\\\infty\\$','\\${\\\\infty}^{0}\\$','\\$\\\\simplify{(sinh({-pa}*x))^({nb}/x)}\\$','\\$0\\\\cdot \\\\infty\\$','\\$\\\\simplify{(sinh({-pa}*x))^({nb}/x)=e^(({nb}/x)*ln(sinh({-pa}*x)))}\\$','\\$\\\\frac{\\\\var{nb*infinity}}{-\\\\infty}\\$','\\$\\\\simplify{(sinh({-pa}*x))^({nb}/x)=e^(({nb}/x)*ln(sinh({-pa}*x)))=e^({nb}ln(sinh({-pa}*x))/x)}\\$','\\$\\\\simplify{e^(({-pa*nb}cosh({-pa}*x))/(sinh({-pa}*x)))=e^({-pa*nb}coth({-pa}*x))}\\$',e^(-pa*nb),'\\$e^{\\\\var{-pa*nb}}\\$']\n\n \n]", "templateType": "anything"}, "pa": {"name": "pa", "group": "Ungrouped variables", "description": "", "definition": "pl[0]", "templateType": "anything"}, "nb": {"name": "nb", "group": "Ungrouped variables", "description": "", "definition": "nl[1]", "templateType": "anything"}, "ne": {"name": "ne", "group": "Ungrouped variables", "description": "", "definition": "nl[4]", "templateType": "anything"}, "pb": {"name": "pb", "group": "Ungrouped variables", "description": "", "definition": "pl[1]", "templateType": "anything"}, "choice1": {"name": "choice1", "group": "Ungrouped variables", "description": "", "definition": "random(zero_zero,inf_zero,one_inf)", "templateType": "anything"}, "na": {"name": "na", "group": "Ungrouped variables", "description": "", "definition": "nl[0]", "templateType": "anything"}, "choice2": {"name": "choice2", "group": "Ungrouped variables", "description": "", "definition": "random(choice1)", "templateType": "anything"}, "pe": {"name": "pe", "group": "Ungrouped variables", "description": "", "definition": "pl[4]", "templateType": "anything"}, "nL": {"name": "nL", "group": "Ungrouped variables", "description": "", "definition": "shuffle(-12..12 except 0)[0..5]", "templateType": "anything"}, "pL": {"name": "pL", "group": "Ungrouped variables", "description": "", "definition": "shuffle(2..12)[0..5]", "templateType": "anything"}, "pd": {"name": "pd", "group": "Ungrouped variables", "description": "", "definition": "pl[3]", "templateType": "anything"}, "nc": {"name": "nc", "group": "Ungrouped variables", "description": "", "definition": "nl[2]", "templateType": "anything"}, "nd": {"name": "nd", "group": "Ungrouped variables", "description": "", "definition": "nl[3]", "templateType": "anything"}, "alt1": {"name": "alt1", "group": "Ungrouped variables", "description": "", "definition": "if(mod(pa,4)=0 or mod(pa,4)=1,1,-1)", "templateType": "anything"}}, "extensions": [], "ungrouped_variables": ["pL", "pa", "pb", "pd", "pe", "nL", "na", "nb", "ne", "nc", "nd", "alt1", "zero_zero", "inf_zero", "one_inf", "choice1", "choice2"], "variablesTest": {"condition": "", "maxRuns": "165"}, "functions": {}, "preamble": {"css": "", "js": ""}, "parts": [{"marks": 0, "showCorrectAnswer": true, "variableReplacements": [], "extendBaseMarkingAlgorithm": true, "gaps": [{"marks": 1, "showCorrectAnswer": false, "variableReplacements": [], "extendBaseMarkingAlgorithm": true, "failureRate": 1, "unitTests": [], "expectedVariableNames": [], "checkingType": "absdiff", "showPreview": true, "variableReplacementStrategy": "originalfirst", "vsetRangePoints": 5, "showFeedbackIcon": true, "checkingAccuracy": "0.0000000001", "type": "jme", "customMarkingAlgorithm": "", "vsetRange": [0, 1], "answer": "{{choice2}[-2]}", "scripts": {}, "checkVariableNames": false}], "variableReplacementStrategy": "originalfirst", "showFeedbackIcon": true, "type": "gapfill", "customMarkingAlgorithm": "", "unitTests": [], "scripts": {}, "prompt": "As {choice2[0]}, {choice2[2]} approaches [[0]]
\nNote: infinity is simply entered by typing infinity
", "sortAnswers": false}], "variable_groups": [], "statement": "This question is about limits of indeterminate forms.
", "advice": "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]}. We can use a standard trick to write this expression to involve an indeterminate form of type {choice2[3]} in the exponent
\n{choice2[4]}
\nAs usual, to deal with an indeterminate form of type {choice2[3]} we rewrite the product as a quotient, in this case, to involve an indeterminate form of type {choice2[5]} as such
\n{choice2[6]}
\n\n\n\nNow we apply L'Hospital's rule to the exponent, which means we now need to determine what {choice2[7]} 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 {choice2[-1]}.
\n", "type": "question", "contributors": [{"name": "Ben Brawn", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/605/"}, {"name": "JPO AddMath", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2346/"}]}]}], "contributors": [{"name": "Ben Brawn", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/605/"}, {"name": "JPO AddMath", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2346/"}]}