The answer dosen't match the actual graph











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$$lim_{xto 0^+}(tan2x)^x$$



When I solve it by first converting it indeterminant form, and then using L Hopital rule two time i.e. differentiating numerator and denominator two times. I finally get the answer as $0$.



However, after looking at the graph of the function. As it approaches $0$, the function becomes $1$, not $0$.



Is my question solving wrong or I have any misconceptions?






limits







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    L Hopital rule is for specific cases of limits of the form f(x)/g(x), and does not apply to this case. In fact, a similar answer was provided to you here: math.stackexchange.com/questions/2997865/…
    – NoChance
    6 hours ago

















up vote
0
down vote

favorite












$$lim_{xto 0^+}(tan2x)^x$$



When I solve it by first converting it indeterminant form, and then using L Hopital rule two time i.e. differentiating numerator and denominator two times. I finally get the answer as $0$.



However, after looking at the graph of the function. As it approaches $0$, the function becomes $1$, not $0$.



Is my question solving wrong or I have any misconceptions?






limits







share|cite|improve this question









New contributor




Amogh Joshi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • 1




    L Hopital rule is for specific cases of limits of the form f(x)/g(x), and does not apply to this case. In fact, a similar answer was provided to you here: math.stackexchange.com/questions/2997865/…
    – NoChance
    6 hours ago















up vote
0
down vote

favorite









up vote
0
down vote

favorite











$$lim_{xto 0^+}(tan2x)^x$$



When I solve it by first converting it indeterminant form, and then using L Hopital rule two time i.e. differentiating numerator and denominator two times. I finally get the answer as $0$.



However, after looking at the graph of the function. As it approaches $0$, the function becomes $1$, not $0$.



Is my question solving wrong or I have any misconceptions?






limits







share|cite|improve this question









New contributor




Amogh Joshi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











$$lim_{xto 0^+}(tan2x)^x$$



When I solve it by first converting it indeterminant form, and then using L Hopital rule two time i.e. differentiating numerator and denominator two times. I finally get the answer as $0$.



However, after looking at the graph of the function. As it approaches $0$, the function becomes $1$, not $0$.



Is my question solving wrong or I have any misconceptions?






limits




limits






share|cite|improve this question









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Amogh Joshi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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share|cite|improve this question









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edited 7 hours ago









Tianlalu

2,180631




2,180631






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asked 7 hours ago









Amogh Joshi

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183




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Amogh Joshi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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Check out our Code of Conduct.








  • 1




    L Hopital rule is for specific cases of limits of the form f(x)/g(x), and does not apply to this case. In fact, a similar answer was provided to you here: math.stackexchange.com/questions/2997865/…
    – NoChance
    6 hours ago
















  • 1




    L Hopital rule is for specific cases of limits of the form f(x)/g(x), and does not apply to this case. In fact, a similar answer was provided to you here: math.stackexchange.com/questions/2997865/…
    – NoChance
    6 hours ago










1




1




L Hopital rule is for specific cases of limits of the form f(x)/g(x), and does not apply to this case. In fact, a similar answer was provided to you here: math.stackexchange.com/questions/2997865/…
– NoChance
6 hours ago






L Hopital rule is for specific cases of limits of the form f(x)/g(x), and does not apply to this case. In fact, a similar answer was provided to you here: math.stackexchange.com/questions/2997865/…
– NoChance
6 hours ago












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$$lim_{xto 0^+}(tan2x)^x=lim_{xto 0^+}e^{xlog(tan2x)}=e^0=1$$






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    $$lim_{xto 0^+}(tan2x)^x=lim_{xto 0^+}e^{xlog(tan2x)}=e^0=1$$






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      $$lim_{xto 0^+}(tan2x)^x=lim_{xto 0^+}e^{xlog(tan2x)}=e^0=1$$






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        $$lim_{xto 0^+}(tan2x)^x=lim_{xto 0^+}e^{xlog(tan2x)}=e^0=1$$






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        $$lim_{xto 0^+}(tan2x)^x=lim_{xto 0^+}e^{xlog(tan2x)}=e^0=1$$







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        answered 6 hours ago









        user376343

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