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$\lim \limits_{h \to 0} \cfrac{\ln \left(\cfrac{1}{e^2}+h \right)+2}{h}$ is the same as:

$f'(e)$ where $f(x)=\cfrac{1}{x^2}$

$f'(e)$ where $f(x)=-\ln \left(\cfrac{1}{x^2} \right)$

$f'(2)$ where $f(x)=\ln \left(\cfrac{2x}{e} \right)$

$f'\left(\cfrac{1}{e^2} \right)$ where $f(x)= \ln x$