Question

Please find the result !​

Answer 1

Answer:

$\displaystyle - \frac{1}{2}$

Step-by-step explanation:

we would like to compute the following limit:

$\displaystyle \lim _{x \to 0} \left( \frac{1}{ \ln(x + \sqrt{ {x}^{2} + 1} ) } - \frac{1}{ \ln(x + 1) } \right)$

if we substitute 0 directly we would end up with:

$\displaystyle\frac{1}{0} - \frac{1}{0}$

which is an indeterminate form! therefore we need an alternate way to compute the limit to do so simplify the expression and that yields:

$\displaystyle \lim _{x \to 0} \left( \frac{ \ln(x + 1) - \ln(x + \sqrt{ {x}^{2} + 1 } }{ \ln(x + \sqrt{ {x}^{2} + 1} ) \ln(x + 1) } \right)$

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

$\rm \displaystyle \lim _{x \to c} \left( \frac{f(x)}{g(x)} \right) = \lim _{x \to c} \left( \frac{f'(x)}{g'(x)} \right)$

thus apply L'hopital rule which yields:

$\displaystyle \lim _{x \to 0} \left( \frac{ \dfrac{d}{dx} \ln(x + 1) - \ln(x + \sqrt{ {x}^{2} + 1 } }{ \dfrac{d}{dx} \ln(x + \sqrt{ {x}^{2} + 1} ) \ln(x + 1) } \right)$

use difference and Product derivation rule to differentiate the numerator and the denominator respectively which yields:

$\displaystyle \lim _{x \to 0} \left( \frac{ \frac{1}{x + 1} - \frac{1}{ \sqrt{x + 1} } }{ \frac{ \ln(x + 1)}{ \sqrt{ {x}^{2} + 1 } } + \frac{ \ln(x + \sqrt{x ^{2} + 1 } }{x + 1} } \right)$

simplify which yields:

$\displaystyle \lim _{x \to 0} \left( \frac{ \sqrt{ {x}^{2} + 1 } - x - 1 }{ (x + 1)\ln(x + 1 ) + \sqrt{ {x}^{2} + 1} \ln( x + \sqrt{ {x }^{2} + 1} ) } \right)$

unfortunately! it's still an indeterminate form if we substitute 0 for x therefore apply L'hopital rule once again which yields:

$\displaystyle \lim _{x \to 0} \left( \frac{ \dfrac{d}{dx} \sqrt{ {x}^{2} + 1 } - x - 1 }{ \dfrac{d}{dx} (x + 1)\ln(x + 1 ) + \sqrt{ {x}^{2} + 1} \ln( x + \sqrt{ {x }^{2} + 1} ) } \right)$

use difference and sum derivation rule to differentiate the numerator and the denominator respectively and that is yields:

$\displaystyle \lim _{x \to 0} \left( \frac{ \frac{x}{ \sqrt{ {x}^{2} + 1 } } - 1}{ \ln(x + 1) + 2 + \frac{x \ln(x + \sqrt{ {x}^{2} + 1 } ) }{ \sqrt{ {x}^{2} + 1 } } } \right)$

thank god! now it's not an indeterminate form if we substitute 0 for x thus do so which yields:

$\displaystyle \frac{ \frac{0}{ \sqrt{ {0}^{2} + 1 } } - 1}{ \ln(0 + 1) + 2 + \frac{0 \ln(0 + \sqrt{ {0}^{2} + 1 } ) }{ \sqrt{ {0}^{2} + 1 } } }$

simplify which yields:

$\displaystyle - \frac{1}{2}$

finally, we are done!

Answer 2

9514 1404 393

Answer:

  -1/2

Step-by-step explanation:

Evaluating the expression directly at x=0 gives ...

  $\dfrac{1}{\ln(\sqrt{1})}-\dfrac{1}{\ln(1)}=\dfrac{1}{0}-\dfrac{1}{0}\qquad\text{an indeterminate form}$

Using the linear approximations of the log and root functions, we can put this in a form that can be evaluated at x=0.

The approximations of interest are ...

  $\ln(x+1)\approx x\quad\text{for x near 0}\\\\\sqrt{x+1}\approx \dfrac{x}{2}+1\quad\text{for x near 0}$

__

Then as x nears zero, the limit we seek is reasonably approximated by the limit ...

  $\displaystyle\lim_{x\to0}\left(\dfrac{1}{x+\dfrac{x^2}{2}}-\dfrac{1}{x}\right)=\lim_{x\to0}\left(\dfrac{x-(x+\dfrac{x^2}{2})}{x(x+\dfrac{x^2}{2})}\right)\\\\=\lim_{x\to0}\dfrac{-\dfrac{x^2}{2}}{x^2(1+\dfrac{x}{2})}=\lim_{x\to0}\dfrac{-1}{2+x}=\boxed{-\dfrac{1}{2}}$

_____

I find a graphing calculator can often give a good clue as to the limit of a function.