Normed Spaces-proving of proposition

Question

Following proposition is given in the textbook

Let X be a normed space over K. Let x$_n$ $\in$ X for n=1,2,3...., and $\lambda$$_n$ $\in$ K, for n=1,2,3..... and $\lambda$ $\in$ K. Suppose that x$_n$ $\rightarrow$x and $\lambda$$_n$ $\rightarrow$$\lambda$ then

1) x is unique

2){x$_n$} is bounded

3) $\left\|x_n\right\|$ $\rightarrow$ $\left\|x\right\|$ as n $\rightarrow$$\infty$

4) $\lambda$x$_n$ $\rightarrow$$\lambda$x as n $\rightarrow$ $\infty$

I have understood the proof for part 1) and 2). However I am unable to understand proofs for 3) and 4). My doubts are as under:

For 3) the proof says by triangle inequality |$\left\|x\right\|$ - $\left\|y\right\|$| $\leq$$\left\|x-y\right\|$ for each x, y $\in$ X. If {x$_n$} is a sequence in X such that x$_n$ $\rightarrow$x, then from the above inequality we have |$\left\|x_n\right\|$ - $\left\|x\right\|$| $\leq$ $\left\|x_n-x\right\|$ and the desired result follows.

I have not understood the above

For 4) textbook says this part follows from $\left\|\lambda x_n-\lambda x\right\|$ $\leq$ |$\lambda$| $\left\|x_n-x\right\|$

I have not understood the above

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Answer 1

If $(a_n)$ and $(b_n)$ are real sequences with $|a_n| \le b_n$ for all $n$ and $b_n \to 0$, then $a_n \to 0$.