We have a sequence $(x_k)$ in which $\lim(x_k)= x$ and $x > 0$. We also have that $\exists k_0$ such that $y_k > 0 \space \forall k \geq k_0$, then show that $\liminf(x_k y_k) = x \liminf(y_k)$.

I'm learning analysis and have this question, but I'm pretty lost on what to do. I think that we can tell that $b_k$ is bounded, but we don't have that it converges, so the Algebraic Limit Theorems don't apply.

  • 1
    $\begingroup$ Welcome to Mathematics SE. Take a tour. You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. $\endgroup$ Commented Jul 9 at 23:24
  • $\begingroup$ just wondering, what do you mean by $liminf$ Do you mean $\lim \inf$ $\endgroup$
    – GSmith
    Commented Jul 9 at 23:25
  • $\begingroup$ Can you show that if $A$ is the set of all partial limits of $(x_ky_k)_{k\in\mathbb{N}}$ and $\alpha \in A$ then $x\lim_{k\to\infty}inf(y_k) \leq \alpha$? $\endgroup$
    – X4J
    Commented Jul 10 at 2:23
  • $\begingroup$ Take advantage of the (actually unnecessary) hypothesis $y_k>0$: take the logarithms. Prove first that if a real sequence $(a_n)$ converges then for any real sequence $(b_n)$,$$\liminf(a_n+b_n)=\lim a_n+\liminf b_n.$$ $\endgroup$ Commented Jul 10 at 5:15
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    $\begingroup$ Please edit your post to include some minimal work. To begin with, write down your preferred definition of $\liminf$. $\endgroup$ Commented Jul 10 at 5:22


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