# The answer dosen't match the actual graph

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

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?

• 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 Nov 14 '18 at 8:29

$$\lim_{x\to 0^+}(\tan2x)^x=\lim_{x\to 0^+}e^{x\log(\tan2x)}=e^0=1$$