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f(x)=((39-12x)/(x-2))^(-12)−6−5−4−3−2−10123456−3-2.5−2-1.5−1-0.500.511.522.53xy

Exercise

lim⁡x→∞(39−12xx−2)−12\lim_{x\to\infty}\left(\frac{39-12x}{x-2}\right)^{-12}

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Final answer to the exercise

1(−12)12\frac{1}{{\left(-12\right)}^{12}}

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limx→∞(39−12xx−2 )−12
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