Calculating limits

Nov 2019
3
0
Durham
Hi there. I am currently struggling to complete both 39 b. I know the answer to b is e^2, but following taking the 2n down from the index after converting it into a logarithm expression, I get an extra e tacked on the end?
Edit: figured out c
 

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Plato

MHF Helper
Aug 2006
22,474
8,643
Part b) can be written as \(\displaystyle {\left[ {{{\left( {1 + \frac{1}{{n + 1}}} \right)}^n}} \right]^2}\)
 

Prove It

MHF Helper
Aug 2008
12,896
5,001
$\displaystyle \begin{align*} \lim_{n \to \infty}{ \left( n^2 + 2 \right) ^{\frac{1}{n}} } &= \lim_{n \to \infty} \mathrm{e}^{\ln{ \left[ \left( n^2 + 2 \right) ^{\frac{1}{n}} \right] }} \\ &= \mathrm{e}^{ \lim_{n \to \infty} \left\{ \ln{ \left[ \left( n^2 + 2 \right) ^{\frac{1}{n}} \right] } \right\} } \\ &= \mathrm{e}^{ \lim_{n \to \infty} \left[ \frac{1}{n}\,\ln{ \left( n^2 + 2 \right) } \right] } \\ &= \mathrm{e}^{ \lim_{n \to \infty} \left[ \frac{\ln{\left( n^2 + 2 \right) }}{n} \right] } \\ &= \mathrm{e}^{ \lim_{n \to \infty} \left( \frac{\frac{2n}{n^2 + 2}}{1} \right) } \textrm{ by L'Hospital's Rule} \\ &= \mathrm{e}^{ \lim_{n \to \infty} \left( \frac{2n}{n^2 + 1} \right) } \\ &= \mathrm{e}^{ \lim_{n \to \infty} \left( \frac{2}{2n} \right) } \textrm{ by L'Hospital's Rule} \\ &= \mathrm{e}^{ \lim_{n \to \infty} \left( \frac{1}{n} \right) } \\ &= \mathrm{e}^0 \\ &= 1 \end{align*}$
 

Prove It

MHF Helper
Aug 2008
12,896
5,001
$\displaystyle \begin{align*} \lim_{n \to \infty} \left[ n \left( \sqrt{n^2 + 1} - \sqrt{n^2 -1} \right) \right] &= \lim_{n \to \infty} \left[ \frac{ n \left( \sqrt{n^2 + 1} - \sqrt{n^2 - 1} \right) \left( \sqrt{n^2 + 1} + \sqrt{n^2 - 1} \right) }{\sqrt{n^2 + 1} + \sqrt{n^2 - 1} } \right] \\ &= \lim_{n \to \infty} \left[ \frac{n \left[ \left( n^2 + 1 \right) - \left( n^2 - 1 \right) \right] }{\sqrt{n^2 + 1} + \sqrt{n^2 - 1}} \right] \\ &= \lim_{n \to \infty} \left( \frac{2n}{\sqrt{n^2 + 1} + \sqrt{n^2 - 1}} \right) \\ &= \lim_{n \to \infty} \left[ \frac{2n \cdot \frac{1}{n}}{\left( \sqrt{n^2 + 1} + \sqrt{n^2 - 1} \right) \cdot \frac{1}{n}} \right] \\ &= \lim_{n \to \infty} \left( \frac{2}{\sqrt{\frac{n^2 + 1}{n^2}} + \sqrt{\frac{n^2 - 1}{n^2}}} \right) \\ &= \lim_{n \to \infty} \left( \frac{2}{\sqrt{1 + \frac{1}{n^2}} + \sqrt{1 - \frac{1}{n^2}}} \right) \\ &= \frac{2}{\sqrt{1 + 0} + \sqrt{1 - 0}} \\ &= \frac{2}{1 + 1} \\ &= 1 \end{align*}$