painter555ui8n
2022-04-17
Answered

What is the best method to solve the limit $\underset{x\to \mathrm{\infty}}{lim}{(1+\mathrm{sin}\frac{2}{{x}^{2}})}^{{x}^{2}}$ ?

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madking711kzcb

Answered 2022-04-18
Author has **9** answers

Hint:

$\underset{x\to \mathrm{\infty}}{lim}{(1+\mathrm{sin}\frac{2}{{x}^{2}})}^{{x}^{2}}={\left(\underset{x\to \mathrm{\infty}}{lim}{(1+\mathrm{sin}\frac{2}{{x}^{2}})}^{\frac{1}{\mathrm{sin}\left(\frac{2}{{x}^{2}}\right)}}\right)}^{2\underset{x\to \mathrm{\infty}}{lim}\frac{\mathrm{sin}\frac{2}{{x}^{2}}}{\frac{2}{{x}^{2}}}}$

Now for the exponent set$\frac{2}{{x}^{2}}=h\Rightarrow h\to {0}^{+}$

Now for the exponent set

chabinka61jx

Answered 2022-04-19
Author has **12** answers

For questions like this one, I've found asymptotic calculations to be handy. For example:

${(1+\mathrm{sin}\frac{2}{{x}^{2}})}^{{x}^{2}}={e}^{{x}^{2}\mathrm{ln}(1+\mathrm{sin}\frac{2}{{x}^{2}})}={e}^{{x}^{2}(\mathrm{sin}\frac{2}{{x}^{2}}+o\left(\mathrm{sin}\frac{2}{{x}^{2}}\right))}={e}^{{x}^{2}(\frac{2}{{x}^{2}}+o\left(\frac{2}{{x}^{2}}\right)+o\left(\frac{2}{{x}^{2}}\right))}={e}^{2+o\left(1\right)}\to {e}^{2}(x\to \mathrm{\infty})$

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The figure is something like:

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Antiderivative of a polynomial fraction

I am trying to solve this problem and I can't seem to understand where I am going wrong. My understanding of antiderivatives is this formula:

$\int {x}^{n}\mathrm{d}x=\frac{{x}^{n+1}}{n+1}+c$

The problem I am trying to solve is:

$f(x)=\frac{{x}^{5}-{x}^{3}+8x}{{x}^{4}}$

I think there are two possible parts that I am getting wrong, I started the problem by separating them into individual fractions and solving each fraction at a time. I think I get it wrong at either the ln(x) or $\frac{8{x}^{-2}}{-2}$. Here is my attempt: $f(x)=\frac{{x}^{5}}{{x}^{4}}-\frac{{x}^{3}}{{x}^{4}}+\frac{8x}{{x}^{4}}=x-\frac{1}{x}+\frac{8}{{x}^{3}}$

Then from here I get the antiderivative:

$F(x)=\frac{{x}^{2}}{2}-\mathrm{ln}(x)+\frac{8{x}^{-2}}{-2}+c=\frac{{x}^{2}}{2}-\mathrm{ln}(x)-\frac{4}{{x}^{2}}+c$

I keep getting the wrong answer when I plug it in, am I missing some fundamental piece of information about antiderivatives which is causing me to mess up the ln(x) or is my algebra wrong and I mess up around $\frac{8{x}^{-2}}{-2}$?

I am trying to solve this problem and I can't seem to understand where I am going wrong. My understanding of antiderivatives is this formula:

$\int {x}^{n}\mathrm{d}x=\frac{{x}^{n+1}}{n+1}+c$

The problem I am trying to solve is:

$f(x)=\frac{{x}^{5}-{x}^{3}+8x}{{x}^{4}}$

I think there are two possible parts that I am getting wrong, I started the problem by separating them into individual fractions and solving each fraction at a time. I think I get it wrong at either the ln(x) or $\frac{8{x}^{-2}}{-2}$. Here is my attempt: $f(x)=\frac{{x}^{5}}{{x}^{4}}-\frac{{x}^{3}}{{x}^{4}}+\frac{8x}{{x}^{4}}=x-\frac{1}{x}+\frac{8}{{x}^{3}}$

Then from here I get the antiderivative:

$F(x)=\frac{{x}^{2}}{2}-\mathrm{ln}(x)+\frac{8{x}^{-2}}{-2}+c=\frac{{x}^{2}}{2}-\mathrm{ln}(x)-\frac{4}{{x}^{2}}+c$

I keep getting the wrong answer when I plug it in, am I missing some fundamental piece of information about antiderivatives which is causing me to mess up the ln(x) or is my algebra wrong and I mess up around $\frac{8{x}^{-2}}{-2}$?

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$z={x}^{2}+{y}^{2}$