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limx→a(2−ax)tanπx2a=en\large \lim_{x \to a} \bigg( 2 - \frac{a}{x} \bigg)^{\tan \frac{\pi x}{2a}} = e^{n}x→alim(2−xa)tan2aπx=en
Find approximate value of nnn.
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limn→∞(a+bn−1a)n \large \displaystyle \lim_{n \to \infty} \left(\dfrac{a+\sqrt[n]{b}-1}{a}\right)^{n} n→∞lim(aa+nb−1)n
Calculate the limit above in terms of aaa and bbb, where aaa and bbb are constants with b≥0b\geq0 b≥0.
Let f:(1,∞)→(0,∞)f:(1,\infty) \rightarrow (0,\infty)f:(1,∞)→(0,∞) be a continuous decreasing function with limx→∞f(4x)f(8x)= 1 \large \lim_{x\to\infty} \dfrac{f(4x)}{f(8x)} = \, 1 x→∞limf(8x)f(4x)=1
Then
limx→∞f(6x)f(8x)= ? \large \lim_{x\to\infty} \dfrac{f(6x)}{f(8x)} = \, ?x→∞limf(8x)f(6x)=?
Let {xn}\{x_n\}{xn} be a sequence such that x1=1, xnxn+1=2nx_1=1,\ x_nx_{n+1}=2nx1=1, xnxn+1=2n for n≥1n\ge 1n≥1.
Find limn→∞∣xn+1−xn∣n.\displaystyle \lim_{n\to\infty} \dfrac{|x_{n+1}-x_n|}{\sqrt{n}}.n→∞limn∣xn+1−xn∣.
Compute limn→∞2016(12015+22015+⋯+n2015)−n20162016(12014+22014+⋯+n2014). \large \lim_{ n\to{\infty}}{\dfrac{ 2016(1^{2015}+2^{2015}+ \cdots +n^{2015}) - n^{2016}}{2016(1^{2014}+2^{2014}+\cdots+n^{2014})}}.n→∞lim2016(12014+22014+⋯+n2014)2016(12015+22015+⋯+n2015)−n2016.
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