The sequence is defined by recurrence: ${x}_{1}=1$, ${x}_{2}=2$

${x}_{n+1}=\frac{1}{2}({x}_{n}+{x}_{n-1})$

${x}_{n+1}=\frac{1}{2}({x}_{n}+{x}_{n-1})$

skylsn
2022-06-26
Answered

The sequence is defined by recurrence: ${x}_{1}=1$, ${x}_{2}=2$

${x}_{n+1}=\frac{1}{2}({x}_{n}+{x}_{n-1})$

${x}_{n+1}=\frac{1}{2}({x}_{n}+{x}_{n-1})$

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asked 2022-07-10

I have one question about limits: it is required to find the limit $\underset{n\to \mathrm{\infty}}{lim}\sqrt[n]{{3}^{n}+{2}^{n}}$

asked 2022-07-09

We have the following limits:

$\underset{x\to \frac{\pi}{4}}{lim}{\displaystyle \frac{1-{\mathrm{tan}}^{2}x}{\mathrm{cos}2x}}$ and $\underset{\theta \to 1}{lim}={\displaystyle \frac{\theta -1+\mathrm{sin}({\theta}^{2}-1)}{{\theta}^{2}-1}}$

Using L'Hôpital I obtained that the second limit is $\theta ={\displaystyle \frac{3}{2}}$. For the first limit I was able to rewrite it to $\underset{x\to \frac{\pi}{4}}{lim}(1-{\mathrm{tan}}^{2}(x))\mathrm{sec}(2x)$, but I don't really know what to do with that.

$\underset{x\to \frac{\pi}{4}}{lim}{\displaystyle \frac{1-{\mathrm{tan}}^{2}x}{\mathrm{cos}2x}}$ and $\underset{\theta \to 1}{lim}={\displaystyle \frac{\theta -1+\mathrm{sin}({\theta}^{2}-1)}{{\theta}^{2}-1}}$

Using L'Hôpital I obtained that the second limit is $\theta ={\displaystyle \frac{3}{2}}$. For the first limit I was able to rewrite it to $\underset{x\to \frac{\pi}{4}}{lim}(1-{\mathrm{tan}}^{2}(x))\mathrm{sec}(2x)$, but I don't really know what to do with that.

asked 2022-06-01

Prove resemblance between $\sum _{n=1}^{\mathrm{\infty}}\mathrm{sin}(2nx)$ and $\mathrm{cot}x$

asked 2022-07-08

I am having trouble finding the limit for the following function:

$\underset{x\to 0}{lim}f(x)=\frac{\mathrm{tan}3x}{\mathrm{tan}2x}$

I do not want to use L'Hospital's rule.

I know that $\frac{\mathrm{tan}3x}{\mathrm{tan}2x}=\frac{\mathrm{sin}3x\mathrm{cos}2x}{\mathrm{sin}2x\mathrm{cos}3x}$, but that does not help me.

Can you maybe give me a hint?

$\underset{x\to 0}{lim}f(x)=\frac{\mathrm{tan}3x}{\mathrm{tan}2x}$

I do not want to use L'Hospital's rule.

I know that $\frac{\mathrm{tan}3x}{\mathrm{tan}2x}=\frac{\mathrm{sin}3x\mathrm{cos}2x}{\mathrm{sin}2x\mathrm{cos}3x}$, but that does not help me.

Can you maybe give me a hint?

asked 2022-07-12

Evaluating $\underset{x\to 0}{lim}\phantom{\rule{thickmathspace}{0ex}}\frac{2\mathrm{cos}(a+x)-\mathrm{cos}(a+2x)-\mathrm{cos}(a)}{{x}^{2}}$

asked 2022-06-22

Geometric (Trigonometric) inequality $\frac{(a+b+c{)}^{3}}{3abc}\le 1+\frac{4R}{r}$

asked 2022-06-13

Calculating $\underset{x\to 0}{lim}\frac{\mathrm{sec}(x)-1}{{x}^{2}\mathrm{sec}(x)}$

The first thing to do, as I was taught, was to rewrite this in terms of sine and cosine. Since $\mathrm{sec}(x)=\frac{1}{\mathrm{cos}(x)}$ we have

$\frac{\frac{1}{\mathrm{cos}(x)}-1}{{x}^{2}\frac{1}{\mathrm{cos}(x)}}$

And that is

$\frac{\frac{1-\mathrm{cos}(x)}{\mathrm{cos}(x)}}{\frac{{x}^{2}}{\mathrm{cos}(x)}}$

Then

$\frac{(1-\mathrm{cos}(x))\cdot \mathrm{cos}(x)}{\mathrm{cos}(x)\cdot {x}^{2}}$

And

$\frac{(1-\mathrm{cos}(x))}{{x}^{2}}$

What was wrong with my procedure?

The first thing to do, as I was taught, was to rewrite this in terms of sine and cosine. Since $\mathrm{sec}(x)=\frac{1}{\mathrm{cos}(x)}$ we have

$\frac{\frac{1}{\mathrm{cos}(x)}-1}{{x}^{2}\frac{1}{\mathrm{cos}(x)}}$

And that is

$\frac{\frac{1-\mathrm{cos}(x)}{\mathrm{cos}(x)}}{\frac{{x}^{2}}{\mathrm{cos}(x)}}$

Then

$\frac{(1-\mathrm{cos}(x))\cdot \mathrm{cos}(x)}{\mathrm{cos}(x)\cdot {x}^{2}}$

And

$\frac{(1-\mathrm{cos}(x))}{{x}^{2}}$

What was wrong with my procedure?