Given positive sequence $a_n$ where $\lim _{n\to \infty} a_n = L, L >0$, prove using the limit definition that $$\displaystyle{\lim _{n\to \infty}}\frac{1}{a_n} = \frac{1}{L}.$$

My thoughts:

How can I do this if I don't know how $a_n$ is defined? I can use the given limit to get the range of $a_n$ in terms of $L$, but I lack the direction to complete the proof.


You don't know the expression of $a_n$ though you know the limit of the sequence as $n\to\infty$ is $L$, which gives you useful information:

  • (1) the sequence $(a_n)$ is bounded;
  • (2) $|a_n-L|$ is small when $n$ is large;
  • (3) $a_n$ is away from $0$ when $n$ is large since $L>0$.

Note that $$ \left|\frac{1}{a_n}-\frac{1}{L}\right|=\frac{|a_n-L|}{|a_n|\cdot L}. $$

You want to show that this quantity is small for large $n$. Well, you can make the numerator small by (2). Now use (3) to show that $1/|a_n|$ cannot be too big.


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