Answer:
[tex] \displaystyle - \frac{1}{2} [/tex]
Step-by-step explanation:
we would like to compute the following limit:
[tex] \displaystyle \lim _{x \to 0} \left( \frac{1}{ \ln(x + \sqrt{ {x}^{2} + 1} ) } - \frac{1}{ \ln(x + 1) } \right) [/tex]
if we substitute 0 directly we would end up with:
[tex] \displaystyle\frac{1}{0} - \frac{1}{0} [/tex]
which is an indeterminate form! therefore we need an alternate way to compute the limit to do so simplify the expression and that yields:
[tex] \displaystyle \lim _{x \to 0} \left( \frac{ \ln(x + 1) - \ln(x + \sqrt{ {x}^{2} + 1 } }{ \ln(x + \sqrt{ {x}^{2} + 1} ) \ln(x + 1) } \right) [/tex]
now notice that after simplifying we ended up with a rational expression in that case to compute the limit we can consider using L'hopital rule which states that
[tex] \rm \displaystyle \lim _{x \to c} \left( \frac{f(x)}{g(x)} \right) = \lim _{x \to c} \left( \frac{f'(x)}{g'(x)} \right) [/tex]
thus apply L'hopital rule which yields:
[tex] \displaystyle \lim _{x \to 0} \left( \frac{ \dfrac{d}{dx} \ln(x + 1) - \ln(x + \sqrt{ {x}^{2} + 1 } }{ \dfrac{d}{dx} \ln(x + \sqrt{ {x}^{2} + 1} ) \ln(x + 1) } \right) [/tex]
use difference and Product derivation rule to differentiate the numerator and the denominator respectively which yields:
[tex] \displaystyle \lim _{x \to 0} \left( \frac{ \frac{1}{x + 1} - \frac{1}{ \sqrt{x + 1} } }{ \frac{ \ln(x + 1)}{ \sqrt{ {x}^{2} + 1 } } + \frac{ \ln(x + \sqrt{x ^{2} + 1 } }{x + 1} } \right) [/tex]
simplify which yields:
[tex] \displaystyle \lim _{x \to 0} \left( \frac{ \sqrt{ {x}^{2} + 1 } - x - 1 }{ (x + 1)\ln(x + 1 ) + \sqrt{ {x}^{2} + 1} \ln( x + \sqrt{ {x }^{2} + 1} ) } \right) [/tex]
unfortunately! it's still an indeterminate form if we substitute 0 for x therefore apply L'hopital rule once again which yields:
[tex] \displaystyle \lim _{x \to 0} \left( \frac{ \dfrac{d}{dx} \sqrt{ {x}^{2} + 1 } - x - 1 }{ \dfrac{d}{dx} (x + 1)\ln(x + 1 ) + \sqrt{ {x}^{2} + 1} \ln( x + \sqrt{ {x }^{2} + 1} ) } \right) [/tex]
use difference and sum derivation rule to differentiate the numerator and the denominator respectively and that is yields:
[tex] \displaystyle \lim _{x \to 0} \left( \frac{ \frac{x}{ \sqrt{ {x}^{2} + 1 } } - 1}{ \ln(x + 1) + 2 + \frac{x \ln(x + \sqrt{ {x}^{2} + 1 } ) }{ \sqrt{ {x}^{2} + 1 } } } \right) [/tex]
thank god! now it's not an indeterminate form if we substitute 0 for x thus do so which yields:
[tex] \displaystyle \frac{ \frac{0}{ \sqrt{ {0}^{2} + 1 } } - 1}{ \ln(0 + 1) + 2 + \frac{0 \ln(0 + \sqrt{ {0}^{2} + 1 } ) }{ \sqrt{ {0}^{2} + 1 } } } [/tex]
simplify which yields:
[tex] \displaystyle - \frac{1}{2} [/tex]
finally, we are done!
