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do's question at Yahoo! Answers regarding the evaluation of a limit

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MarkFL

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Feb 24, 2012
13,775
Here is the question:

How do you find ths prob lim(x to(-)infty)frac(14-13 x)(10+x)+\frac(5 x^2 +14)((11 x-12)^2)?
I have posted a link there to this topic so the OP can see my work.
 
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MarkFL

Administrator
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Feb 24, 2012
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Hello do,

We are given to evaluate:

\(\displaystyle \lim_{x\to-\infty}\left(\frac{14-13x}{10+x}+\frac{5x^2+14}{(11x-12)^2} \right)\)

A property of limits that we can use here is:

\(\displaystyle \lim_{x\to c}\left(f(x)\pm g(x) \right)=\lim_{x\to c}(f(x))\pm\lim_{x\to c}(g(x))\)

And so we may rewrite the limit as:

\(\displaystyle \lim_{x\to-\infty}\left(\frac{5x^2+14}{(11x-12)^2} \right)-\lim_{x\to-\infty}\left(\frac{13x-14}{x+10} \right)\)

Next, consider a function of the form:

\(\displaystyle f(x)=\frac{a_nx^n+a_{n-1}x^{n-1}+\cdots+a_0}{b_nx^n+b_{n-1}x^{n-1}+\cdots+b_0}\)

Now, if we divide each term in the numerator and denominator by $x^n$, we have:

\(\displaystyle f(x)=\frac{a_n+a_{n-1}x^{-1}+\cdots+a_0x^{-n}}{b_n+b_{n-1}x^{-1}+\cdots+b_0x^{-n}}\)

And so, we see:

\(\displaystyle \lim_{x\to\pm\infty}(f(x))=\frac{a_n+0+\cdots+0}{b_n+0+\cdots+0}=\frac{a_n}{b_n}\)

Applying this to the limit at hand, we find:

\(\displaystyle \lim_{x\to-\infty}\left(\frac{5x^2+14}{(11x-12)^2} \right)-\lim_{x\to-\infty}\left(\frac{13x-14}{x+10} \right)=\frac{5}{121}-\frac{13}{1}=-\frac{1568}{121}\)