Proving convergence of sequence using epsilon-N

Question

Prove $\lim_{n\to∞}(256+v_n)^{1/8}=2$ if $|v_n|<1$ and $\lim_{n\to∞}(v_n)=0$

I have attempted this question by using difference of two squares to simplify the nominator to $|v_n|$ and then used inequality arguments and the property of $|v_n|<1$ to further simplify the fraction however don't know where to go from here.

I believe the way in which I'm attempting this question is wrong, and would appreciate any hints/guides to solve this.

Thanks

Answer 1

$$a^8-b^8=(a^4-b^4)(a^4+b^4)=(a^2-b^2)(a^2+b^2)(a^4+b^4)=(a-b)(a+b)(a^2+b^2)(a^4+b^4)$$

That is $$a-b=\frac{a^8-b^8}{(a+b)(a^2+b^2)(a^4+b^4)}$$

$$(256+v_n)^{\frac18}-2=\frac{v_n}{[(256+v_n)^{\frac18}+2][(256+v_n)^{\frac14}+2^2][(256+v_n)^{\frac12}+2^4]}$$

Let's try to bound the denominator:

For example,

We know that $|v_n|<1$, hence $255<256+v_n<257$,

$$255^\frac18<(256+v_n)^\frac18<257^\frac18$$

$$2<255^\frac18+2<(256+v_n)^\frac18+2<257^\frac18+2$$

Similarly, we can find similar bounds for the other two terms and we can write

$$|(256+v_n)^{\frac18}-2|\le\frac{|v_n|}{A}$$

for some positive number $A$.

Now, hopefully an $\epsilon-N$ proof is doable.

Answer 2

Hint: Use the power rule of limits,

https://www.math24.net/properties-limits/

Answer 3

By the product rule, we have

$$\left(\lim_{n\to\infty}a_n\right)^2=\lim_{n\to\infty}a_n^2.$$

Applying it three times,

$$\left(\lim_{n\to\infty}(256+v_n)^{1/8}\right)^8=\lim_{n\to\infty}(256+v_n)=256.$$

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The first identity can be proven with

$$|a_n^2-a^2|=|a_n-a||a_n-a+2a|\le|a_n-a|^2+2a|a_n-a|\le\delta^2+2a\delta.$$