Limit of difference quotient of $\frac{{t}^{4}}{2}+{t}^{3}-\frac{{t}^{2}}{2}+1$

Feinsn
2022-06-05
Answered

Limit of difference quotient of $\frac{{t}^{4}}{2}+{t}^{3}-\frac{{t}^{2}}{2}+1$

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hildiadau0o

Answered 2022-06-06
Author has **21** answers

You made a mistake in that step, the correct is as below

$\underset{\text{}\mathrm{\Delta}t\to 0}{lim}\frac{x(1+\mathrm{\Delta}t)-x(1)}{\mathrm{\Delta}t}=\underset{\mathrm{\Delta}t\to 0}{lim}\frac{\frac{\mathrm{\Delta}{t}^{4}+6\mathrm{\Delta}{t}^{3}+11\mathrm{\Delta}{t}^{2}+8\mathrm{\Delta}t}{2}}{\mathrm{\Delta}t}=\frac{1}{2}\underset{\mathrm{\Delta}t\to 0}{lim}\frac{\mathrm{\Delta}{t}^{3}+6\mathrm{\Delta}{t}^{2}+11\mathrm{\Delta}t+8}{2}=4$

$\underset{\text{}\mathrm{\Delta}t\to 0}{lim}\frac{x(1+\mathrm{\Delta}t)-x(1)}{\mathrm{\Delta}t}=\underset{\mathrm{\Delta}t\to 0}{lim}\frac{\frac{\mathrm{\Delta}{t}^{4}+6\mathrm{\Delta}{t}^{3}+11\mathrm{\Delta}{t}^{2}+8\mathrm{\Delta}t}{2}}{\mathrm{\Delta}t}=\frac{1}{2}\underset{\mathrm{\Delta}t\to 0}{lim}\frac{\mathrm{\Delta}{t}^{3}+6\mathrm{\Delta}{t}^{2}+11\mathrm{\Delta}t+8}{2}=4$

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The figure shows the surface created when the cylinder ${y}^{2}+{Z}^{2}=1$ intersects the cylinder ${x}^{2}+{Z}^{2}=1$. Find the area of this surface.

The figure is something like:

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Antiderivative of $x{e}^{-c{x}^{2}}$

I need to define c in ${\int}_{0}^{\mathrm{\infty}}x{e}^{-c{x}^{2}},$, so that it becomes a probability-mass function (so that it equals 1).

Where do I even begin finding the antiderivative of this? I know the answer will be: $\frac{{e}^{-c{x}^{2}}}{-2c}$.

Trying to use partial integration:

$\int f(x)g(x)dx=f(x)G(x)-\int {f}^{\prime}(x)G(X)$

and picking x as my g(x), and ${e}^{-c{x}^{2}}$ as my f(x) I end up with:

${f}^{\prime}(x)=-2cxf(x)$

$G(X)=\frac{{x}^{2}}{2}$

${e}^{-c{x}^{2}}\frac{{x}^{2}}{2}-{\int}_{0}^{\mathrm{\infty}}-2cxf(x)\frac{{x}^{2}}{2}dx$

Which simplifies to: ${e}^{-c{x}^{2}}\frac{{x}^{2}}{2}-{\int}_{0}^{\mathrm{\infty}}-c{x}^{4}{e}^{-c{x}^{2}}dx$, which I find is just a mess.

I need to define c in ${\int}_{0}^{\mathrm{\infty}}x{e}^{-c{x}^{2}},$, so that it becomes a probability-mass function (so that it equals 1).

Where do I even begin finding the antiderivative of this? I know the answer will be: $\frac{{e}^{-c{x}^{2}}}{-2c}$.

Trying to use partial integration:

$\int f(x)g(x)dx=f(x)G(x)-\int {f}^{\prime}(x)G(X)$

and picking x as my g(x), and ${e}^{-c{x}^{2}}$ as my f(x) I end up with:

${f}^{\prime}(x)=-2cxf(x)$

$G(X)=\frac{{x}^{2}}{2}$

${e}^{-c{x}^{2}}\frac{{x}^{2}}{2}-{\int}_{0}^{\mathrm{\infty}}-2cxf(x)\frac{{x}^{2}}{2}dx$

Which simplifies to: ${e}^{-c{x}^{2}}\frac{{x}^{2}}{2}-{\int}_{0}^{\mathrm{\infty}}-c{x}^{4}{e}^{-c{x}^{2}}dx$, which I find is just a mess.

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Divide numerator and denominator by the highest power of x in the denominator and proceed from there. Find the limits

$\underset{x\to -\mathrm{\infty}}{lim}{\left(\frac{{x}^{2}+x-1}{8{x}^{2}-3}\right)}^{\frac{1}{3}}$

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Find derivatives of the function defined as follows.

$y={e}^{-8x}$

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Find the limit:

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