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

grenivkah3z
2022-07-10
Answered

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

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Isla Klein

Answered 2022-07-11
Author has **12** answers

Notice that

$({3}^{n}+{2}^{n}{)}^{1/n}={({5}^{n}\cdot \frac{{3}^{n}+{2}^{n}}{{5}^{n}})}^{1/n}=5\cdot {[{\left(\frac{3}{5}\right)}^{n}+{\left(\frac{2}{5}\right)}^{n}]}^{1/n}.$

Thus, when you concluded that the second limit is 0 you virtually assumed that

$\underset{n\to \mathrm{\infty}}{lim}{[{\left(\frac{3}{5}\right)}^{n}+{\left(\frac{2}{5}\right)}^{n}]}^{1/n}=0,$

which is false. Your assumption is that since

$\underset{n\to \mathrm{\infty}}{lim}{\left(\frac{3}{5}\right)}^{n}+{\left(\frac{2}{5}\right)}^{n}=0,$

that the former limit must also be 0 However, this is not the case: in the cases where the exponent has limit 0 it is not sufficient for the base to have limit 0 in order for the entire power to have limit 0 The exponent is $1/n,$ and notice that

$\underset{n\to \mathrm{\infty}}{lim}\frac{1}{n}=0.$

As such, your second calculation is simply incorrect. In general, if

$\underset{n\to \mathrm{\infty}}{lim}{a}_{n}=0$

and

$\underset{n\to \mathrm{\infty}}{lim}{b}_{n}=0,$

then

$\underset{n\to \mathrm{\infty}}{lim}{{b}_{n}}^{{a}_{n}}$

can be equal to any nonnegative real number, or it can even not exist. You cannot simply conclude it to be 0.

$({3}^{n}+{2}^{n}{)}^{1/n}={({5}^{n}\cdot \frac{{3}^{n}+{2}^{n}}{{5}^{n}})}^{1/n}=5\cdot {[{\left(\frac{3}{5}\right)}^{n}+{\left(\frac{2}{5}\right)}^{n}]}^{1/n}.$

Thus, when you concluded that the second limit is 0 you virtually assumed that

$\underset{n\to \mathrm{\infty}}{lim}{[{\left(\frac{3}{5}\right)}^{n}+{\left(\frac{2}{5}\right)}^{n}]}^{1/n}=0,$

which is false. Your assumption is that since

$\underset{n\to \mathrm{\infty}}{lim}{\left(\frac{3}{5}\right)}^{n}+{\left(\frac{2}{5}\right)}^{n}=0,$

that the former limit must also be 0 However, this is not the case: in the cases where the exponent has limit 0 it is not sufficient for the base to have limit 0 in order for the entire power to have limit 0 The exponent is $1/n,$ and notice that

$\underset{n\to \mathrm{\infty}}{lim}\frac{1}{n}=0.$

As such, your second calculation is simply incorrect. In general, if

$\underset{n\to \mathrm{\infty}}{lim}{a}_{n}=0$

and

$\underset{n\to \mathrm{\infty}}{lim}{b}_{n}=0,$

then

$\underset{n\to \mathrm{\infty}}{lim}{{b}_{n}}^{{a}_{n}}$

can be equal to any nonnegative real number, or it can even not exist. You cannot simply conclude it to be 0.

babyagelesszj

Answered 2022-07-12
Author has **7** answers

The expression just equals:-

$\underset{n\to \mathrm{\infty}}{lim}3{(1+(\frac{2}{3}{)}^{n})}^{\frac{1}{n}}$

Now this is ${1}^{\mathrm{\infty}}$ form.

You can use this whenever you run into this ${1}^{\mathrm{\infty}}$ form

$\underset{x\to a}{lim}f(x{)}^{g(x)}=\mathrm{exp}(\underset{x\to a}{lim}(f(x)-1)g(x))$

when $\underset{x\to a}{lim}f(x{)}^{g(x)}$ is ${1}^{\mathrm{\infty}}$ form.

So using this we get it as :-

$3\mathrm{exp}(\underset{n\to \mathrm{\infty}}{lim}\frac{(\frac{2}{3}{)}^{n}}{n})=3\mathrm{exp}(0)=3$

In general whenever you have something like

$\underset{n\to \mathrm{\infty}}{lim}{({a}_{1}^{n}+{a}_{2}^{n}+...+{a}_{m}^{n})}^{\frac{1}{n}}$

where ${a}_{i}$ 's are distinct positive real numbers.You can use the above method and immediately write the answer as $\underset{1\le i\le m}{max}{a}_{i}$

That is we are strongly using the fact that ${a}^{n}\to 0$ when $0<a<1$

$\underset{n\to \mathrm{\infty}}{lim}3{(1+(\frac{2}{3}{)}^{n})}^{\frac{1}{n}}$

Now this is ${1}^{\mathrm{\infty}}$ form.

You can use this whenever you run into this ${1}^{\mathrm{\infty}}$ form

$\underset{x\to a}{lim}f(x{)}^{g(x)}=\mathrm{exp}(\underset{x\to a}{lim}(f(x)-1)g(x))$

when $\underset{x\to a}{lim}f(x{)}^{g(x)}$ is ${1}^{\mathrm{\infty}}$ form.

So using this we get it as :-

$3\mathrm{exp}(\underset{n\to \mathrm{\infty}}{lim}\frac{(\frac{2}{3}{)}^{n}}{n})=3\mathrm{exp}(0)=3$

In general whenever you have something like

$\underset{n\to \mathrm{\infty}}{lim}{({a}_{1}^{n}+{a}_{2}^{n}+...+{a}_{m}^{n})}^{\frac{1}{n}}$

where ${a}_{i}$ 's are distinct positive real numbers.You can use the above method and immediately write the answer as $\underset{1\le i\le m}{max}{a}_{i}$

That is we are strongly using the fact that ${a}^{n}\to 0$ when $0<a<1$

asked 2022-05-23

Given is the sequence ${x}_{1}=0,\phantom{\rule{thickmathspace}{0ex}}{x}_{n+1}=\sqrt{2+{x}_{n}}$. Prove:

$\underset{n\to \mathrm{\infty}}{lim}{2}^{n}\sqrt{2-{x}_{n}}=\pi $

Hint:

Use the following formulas:

$\mathrm{cos}\left(\frac{x}{2}\right)=\sqrt{\frac{1+\mathrm{cos}x}{2}}$

$\mathrm{sin}\left(\frac{x}{2}\right)=\sqrt{\frac{1-\mathrm{cos}x}{2}}$

Any idea how to solve this problem?

$\underset{n\to \mathrm{\infty}}{lim}{2}^{n}\sqrt{2-{x}_{n}}=\pi $

Hint:

Use the following formulas:

$\mathrm{cos}\left(\frac{x}{2}\right)=\sqrt{\frac{1+\mathrm{cos}x}{2}}$

$\mathrm{sin}\left(\frac{x}{2}\right)=\sqrt{\frac{1-\mathrm{cos}x}{2}}$

Any idea how to solve this problem?

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$\underset{n\to \mathrm{\infty}}{lim}{u}_{n}$

$\underset{n\to \mathrm{\infty}}{lim}{u}_{n}$

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