MHB Do's question at Yahoo Answers regarding the evaluation of a limit

[MarkFL]

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.

 

[MarkFL]

Hello do,

We are given to evaluate:

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

A property of limits that we can use here is:

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

And so we may rewrite the limit as:

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

Next, consider a function of the form:

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

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

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

And so, we see:

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

Applying this to the limit at hand, we find:

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