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

MATH110 April 2019

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Question 02 (c)
Evaluate $\lim _{x\to \infty }x\ln \left({\frac {x+1}{x}}\right)$

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
Use l'Hospital's rule. This allows us to compute a limit of two functions, where one grows very large (i.e. approaches infinity) and the other approaches zero. See the next hint for more information.

Hint 2
Notice that: $\lim _{x\to \infty }x=\infty$ Also, we have that: $\lim _{x\to \infty }\log({\frac {x+1}{x}})=\log(1)=0$. This justifies our use of L'Hospital's rule.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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.

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