Science:Math Exam Resources/Courses/MATH100/December 2012/Question 04 (c)

MATH100 December 2012

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Question 04 (c)
Short-Answer Questions. Questions 1-4 are short-answer questions. Put your answers in the boxes provided. Simplify your answers as much as possible, and show your work. Each question is worth 3 marks, but not all questions are of equal difficulty. Evaluate $\lim _{x\to \infty }\left(1+{\frac {5}{x}}\right)^{x}$

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 hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Use the fact that $\displaystyle f(x)=e^{\ln f(x)}$ and move the limit to the exponent (valid since exponentials are continuous functions). Then solve the remaining limit using L'Hopital'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. Proceeding as suggested in the hint, we have ${\begin{aligned}\lim _{x\to \infty }\left(1+{\frac {5}{x}}\right)^{x}&=\lim _{x\to \infty }e^{\ln \left(1+{\frac {5}{x}}\right)^{x}}\\&=e^{\lim _{x\to \infty }x\ln \left(1+{\frac {5}{x}}\right)}\\&=e^{\lim _{x\to \infty }{\frac {\ln \left(1+{\frac {5}{x}}\right)}{\tfrac {1}{x}}}}\end{aligned}}$ Where we first use that $\ln(a^{b})=b\ln(a)$, and then multiplied the top and bottom by 1/x. Let us now look at the exponent in more detail. For $x\rightarrow \infty$ the numerator and the denominator both go to 0 (since ln(1) = 0). We hence use L'Hospital's rule for the exponent: $\lim _{x\to \infty }{\frac {\ln \left(1+{\frac {5}{x}}\right)}{\tfrac {1}{x}}}=\lim _{x\to \infty }{\frac {{\frac {1}{1+{\tfrac {5}{x}}}}\cdot {\frac {-5}{x^{2}}}}{-{\tfrac {1}{x^{2}}}}}=\lim _{x\rightarrow \infty }{\frac {5}{1+{\frac {5}{x}}}}$ Where we cancel the $-{\frac {1}{x^{2}}}$ in the second step. The remaining limit is straight forward and evaluates to 5. Hence our final answer is $\lim _{x\to \infty }\left(1+{\frac {5}{x}}\right)^{x}=e^{5}$

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