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[tex]h(x) = x -1 + \frac{1+ ln {}^{2} (x) }{x}[/tex]
[tex]\displaystyle \lim_{x\to0} h(x)= \: ? \\ \displaystyle \lim_{x\to \infty } h(x)= \: ?[/tex]
Apply L'Hôpital's Rule if possible​

Sagot :

Answer:

[tex]\lim_{x\rightarrow +\infty } x-1+\frac{1+ln^{2}x}{x} = + \infty[/tex]

[tex]\lim_{x\rightarrow 0 } x-1+\frac{1+ln^{2}x}{x} = + \infty[/tex]

Step-by-step explanation:

[tex]\lim_{x\rightarrow +\infty } x-1+\frac{1+ln^{2}x}{x}[/tex]

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

=          +∞             +          0

= +∞

[tex]\lim_{x\rightarrow +\infty } x-1+\frac{1+ln^{2}x}{x}[/tex]

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

=            -1             +      +∞  

= +∞  

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