What is the limit of $f (x) = - \frac{1}{2 {(ln x)}^{2}}$ $f(x) = -1/(2(lnx)^2)$ ?

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Answer 1 · 2017-11-16T01:21:32

$lim_(x->+oo) -1/(2(lnx)^2) =0$
$lim_(x->1) -1/(2(lnx)^2) =-oo$

$f(x) = -1/(2(lnx)^2)$

$lnx$ is defined for $x>0 -> f(x)$ is defined for $x>0$
Also, $ln(1) =0 -> f(x)$ is undefined at $(x=1)$

Hence, the domain of $f(x)$ is $(0,+1)uu(+1,+oo)$

Since the question does not specify which limit is sought, I will deal with both possibilities.

$lim_(x->+oo) -1/(2(lnx)^2) -> -1/oo = 0$

Since $lim_(x->1) (lnx)^2 =0$ for $1^-$ and $1^+$

$lim_(x->1) -1/(2(lnx)^2) -> -1/0 =-oo$

We can deduce these results from the graph of $f(x)$ below.

graph{ -1/(2(lnx)^2) [-3.024, 9.463, -3.62, 2.617]}

What is the limit of f(x) = -1/(2(lnx)^2)f(x)=−12(lnx)2f(x) = -1/(2(lnx)^2) ?