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$$\lim_{x\to 0^+}(\tan2x)^x$$

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?

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    $\begingroup$ 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/… $\endgroup$ – NoChance Nov 14 '18 at 8:29
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$$\lim_{x\to 0^+}(\tan2x)^x=\lim_{x\to 0^+}e^{x\log(\tan2x)}=e^0=1$$

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