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do's question at Yahoo! Answers regarding the evaluation of a limit
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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:
\(\displaystyle \lim_{x\to-\infty}\left(\frac{5x^2+14}{(11x-12)^2} \right)-\lim_{x\to-\infty}\left(\frac{13x-14}{x+10} \right)=\frac{5}{121}-\frac{13}{1}=-\frac{1568}{121}\)
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:
\(\displaystyle \lim_{x\to-\infty}\left(\frac{5x^2+14}{(11x-12)^2} \right)-\lim_{x\to-\infty}\left(\frac{13x-14}{x+10} \right)=\frac{5}{121}-\frac{13}{1}=-\frac{1568}{121}\)