Answer:
[tex]\lim_{x \to 0} \frac{2x}{\frac{1}{1+x^{2} } }[/tex] =0
Step-by-step explanation:
[tex]\lim_{x \to 0} \frac{x^{2} }{tan^{-1} (x)}[/tex]
the x value for which tan(x)=0 is 0
we get an indefinite form, so we use l'hopital's rule
(l'hopitals rule: the limit of an indefinite form is equal to the limit of the derivative of the numerator divided by the derivative of the denominator)
[tex]\lim_{x \to 0} \frac{2x}{\frac{1}{1+x^{2} } }[/tex] = 0/1 =0
when we graph the function[tex]\frac{x^{2} }{tan^{-1} (x)}[/tex], the graph seems to pass the point (0,0)