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

Question

`h(x) = x -1 + \frac{1+ ln {}^(2) (x) }{x}`


`\displaystyle \lim_(x\to0) h(x)= \: ? \\ \displaystyle \lim_(x\to \infty ) h(x)= \: ?`
Apply L'Hôpital's Rule if possible​

Answer 1

`\lim_(x\rightarrow +\infty ) x-1+(1+ln^(2)x)/(x) = + \infty`

`\lim_(x\rightarrow 0 ) x-1+(1+ln^(2)x)/(x) = + \infty`

Explanation:

`\lim_(x\rightarrow +\infty ) x-1+(1+ln^(2)x)/(x)`

`= [\lim_(x\rightarrow +\infty ) (x-1)]+[ \lim_(x\rightarrow +\infty ) ((1+ln^(2)x)/(x))]`

= +∞ + 0

= +∞

`\lim_(x\rightarrow +\infty ) x-1+(1+ln^(2)x)/(x)`

`= [\lim_(x\rightarrow 0 ) (x-1)]+[ \lim_(x\rightarrow 0 ) ((1+ln^(2)x)/(x))]`

= -1 + +∞

= +∞