Science:Math Exam Resources/Courses/MATH110/April 2019/Question 02 (c)/Solution 1

Using l'Hospital's rule, we have the following calculations:

$\lim _{x\to \infty }x\ln \left({\frac {x+1}{x}}\right)=\lim _{x\to \infty }{\frac {\ln \left({\frac {x+1}{x}}\right)}{x^{-1}}}=\lim _{x\to \infty }{\frac {{\frac {x}{x+1}}\cdot (-x^{-2})}{-x^{-2}}}=\lim _{x\to \infty }{\frac {x}{x+1}}=1.$

Notice, the first inequality puts the fraction in the correct form to apply l'Hospital's rule. We then apply l'Hospital's rule, which involves differentiating the numerator and denominator. Finally, we evaluate the limit, which gives the final numerical answer.