Question

id="TexFormula1" title="h(x) = x -1 + \frac{1+ ln {}^{2} (x) }{x}" alt="h(x) = x -1 + \frac{1+ ln {}^{2} (x) }{x}" align="absmiddle" class="latex-formula">
$\displaystyle \lim_{x\to0} h(x)= \: ? \\ \displaystyle \lim_{x\to \infty } h(x)= \: ?$
Apply L'Hôpital's Rule if possible​

Answer 1

Answer:

$\lim_{x\rightarrow +\infty } x-1+\frac{1+ln^{2}x}{x} = + \infty$

$\lim_{x\rightarrow 0 } x-1+\frac{1+ln^{2}x}{x} = + \infty$

Step-by-step explanation:

$\lim_{x\rightarrow +\infty } x-1+\frac{1+ln^{2}x}{x}$

$= [\lim_{x\rightarrow +\infty } (x-1)]+[ \lim_{x\rightarrow +\infty } (\frac{1+ln^{2}x}{x})]$

=          +∞             +          0

= +∞

$\lim_{x\rightarrow +\infty } x-1+\frac{1+ln^{2}x}{x}$

$= [\lim_{x\rightarrow 0 } (x-1)]+[ \lim_{x\rightarrow 0 } (\frac{1+ln^{2}x}{x})]$

=            -1             +      +∞  

= +∞