(1) Set up only the partial fraction decomposition of the fraction P(x)/(x + 8)^3(x^2 + 1)^2. (2)...


(1) Set up only the partial fraction decomposition of the fraction {eq}\frac{\displaystyle P(x)}{\displaystyle (x + 8)^3(x^2 + 1)^2} {/eq}

(2) Find the limit: {eq}\displaystyle \lim_{\displaystyle x \to 0} \frac{\displaystyle x + 1 - e^x}{\displaystyle x^2} {/eq}

(3) Find the limit: {eq}\displaystyle \lim_{\displaystyle x \to 0} (3x)^{\frac{\displaystyle 1}{\displaystyle 4}} {/eq}

Partial Fraction and Limits.

In integration, we have familiar with the term partial fraction. here we integrate the rational fraction into the easiest denominator.

Limits explain when the function tends to infinity, minus infinity or to a real limit.

and if we take the limit of the constant function it gives constant only.

The formula is:

{eq}\dfrac{p(x)+q}{(x-a)^2}=\dfrac{A}{x-a}+\dfrac{B}{(x-a)^2} {/eq}

Answer and Explanation:

Part 1.)

We have to set up only the partial fraction of the given function:

{eq}\dfrac{p(x)}{(x+8)^3(x^2+1)^2}\\\\ {/eq}

Hence the partial form decomposition of the given function will be:

{eq}\dfrac{p(x)}{(x+8)^3(x^2+1)^2}=\dfrac{A}{x+8}+\dfrac{B}{(x+8)^2}+\dfrac{C}{(x+8)^3}+\dfrac{Dx+E}{x^2+1}+\dfrac{Fx+G}{(x^2+1)^2}\\\\ {/eq}

Part 2.)

We have to solve the given limits:

{eq}lim_{x\rightarrow 0}\dfrac{x+1-e^x}{x^2}\\\\ {/eq}

Applying L'Hospital rule:

{eq}\dfrac{d}{dx}\left ( \dfrac{x+1-e^x}{x^2} \right )\\\\ =lim_{x\rightarrow 0}\left ( \dfrac{-e^x+1}{2x} \right )\\\\ {/eq}

Again applying L'Hospital rule we get:

{eq}=lim_{x\rightarrow 0}\left ( \dfrac{-e^x}{2} \right )\\\\ =\dfrac{-e^0}{2}\\\\ =\dfrac{-1}{2}\\\\ {/eq}

Part 3.)

{eq}lim_{x\rightarrow 0}(3x)^{\frac{1}{4}}\\\\ {/eq}

Substituting the values of limits {eq}(x\rightarrow 0 ) {/eq} we get:

{eq}=(3\cdot 0)^{\frac{1}{4}}\\\\ =0 {/eq}

Learn more about this topic:

Integration Problems in Calculus: Solutions & Examples

from AP Calculus AB & BC: Homework Help Resource

Chapter 13 / Lesson 13

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