Is it always true? Why?

Algebra Level pending

lim [ f ( x ) + g ( x ) ] = lim f ( x ) + lim g ( x ) \lim \ [f(x)+g(x)]=\lim f(x)+\lim g(x)

Is the above always true?

No Yes

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1 solution

Chew-Seong Cheong
Aug 27, 2016

For example, f ( n ) = k = 1 n ( 1 k 2 1 k ) \displaystyle f(n) = \sum_{k=1}^n \left(\frac 1{k^2} - \frac 1k \right) and g ( n ) = k = 1 n 1 k \displaystyle g(n) = \sum_{k=1}^n \frac 1k . Since lim n k = 1 n 1 k \displaystyle \lim_{n \to \infty} \sum_{k=1}^n \frac 1k does not exist, both lim n f ( x ) \displaystyle \lim_{n \to \infty} f(x) and lim n g ( x ) \displaystyle \lim_{n \to \infty} g(x) do not exist. But the LHS does exist as shown below:

lim n [ f ( n ) + g ( n ) ] = lim n [ k = 1 n ( 1 k 2 1 k ) + k = 1 n 1 k ] = lim n k = 1 n 1 k 2 = ζ ( 2 ) = π 2 6 \begin{aligned} \lim_{n \to \infty} [f(n) + g(n)] & = \lim_{n \to \infty} \left[ \sum_{k=1}^n \left(\frac 1{k^2} - \frac 1k \right) + \sum_{k=1}^n \frac 1k \right] = \lim_{n \to \infty} \sum_{k=1}^n \frac 1{k^2} = \zeta(2) = \frac {\pi^2} 6 \end{aligned}

Not a 3.14../6 but 3.14x3.14/6

Oleg Yovanovich - 3 months, 1 week ago

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Thanks, I have amended it. Key in \pi in LaTex

Chew-Seong Cheong - 3 months, 1 week ago

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