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

  • Thread starter Thread starter MarkFL
  • Start date Start date
  • Tags Tags
    Limit
AI Thread Summary
The limit to evaluate is lim(x→-∞)((14-13x)/(10+x) + (5x^2 + 14)/((11x-12)^2)). By applying the property of limits, the expression can be separated into two parts. The first part simplifies to lim(x→-∞)((5x^2 + 14)/((11x-12)^2)), which approaches 5/121, while the second part, lim(x→-∞)((14-13x)/(10+x)), simplifies to -13. Combining these results gives a final limit of -1568/121. This demonstrates the application of limit properties in evaluating complex rational functions.
MarkFL
Gold Member
MHB
Messages
13,284
Reaction score
12
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.
 
Mathematics news on Phys.org
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:

$$\lim_{x\to-\infty}\left(\frac{5x^2+14}{(11x-12)^2} \right)-\lim_{x\to-\infty}\left(\frac{13x-14}{x+10} \right)=\frac{5}{121}-\frac{13}{1}=-\frac{1568}{121}$$
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Back
Top