lim changes equation when plotting?

[BJWiley23]

I was looking to fresh my trig and calc skills for entrance exams for MS Degrees. I am wondering how I input this first equation into symbolab.com and on the graph populate another equation. The equations equal but I am not sure how to get from the equation entered to the one plotted

Entered:
\lim _{x\to \:3}\left(\frac{5x^2-8x-13}{x^2-5}\right)

then on the graph, the equation that is plotted is as:
y=\left(\frac{12-8x}{x2-5}\right)+5

How you get from \lim _{x\to \:3}\left(\frac{5x^2-8x-13}{x^2-5}\right) to y=\left(\frac{12-8x}{x2-5}\right)+5

Here is the URL to clarify:
https://www.symbolab.com/solver/limit-c ... 7D%5Cright)

Brian

[elenasimons]

a similar problem, did you get a clue on your issue? if so, share it with me

[HallsofIvy]

Well, strictly speaking, you don't go from one to the other because you have a "lim" in one that you don't have in the other! But you go from [tex][tex]\frac{5x^2- 8x- 13}{x^2- 5}[/tex][/tex] to [tex]5+ \frac{12- 8x}{x^2- 5}[/tex] by division! A numerical example is that 17 divides into 385 22 times with remainder 12: [tex]\frac{385}{17}= 22+ \frac{12}{17}[/tex].

The numerator has [tex]5x^2[/tex] while the denominator has only [tex]x^2[/tex]. The "trial divisor" is [tex]\frac{5x^2}{x^2}= 5[/tex]. Then you subtract the entire divisor, [tex]x^2- 5[/tex],times 5, from the dividend, [tex]5x^2- 8x- 13[/tex]:

[tex]5x^2- 8x- 13- 5(x^2+ 0x- 5)= 5x^2- 8x- 13- 5x^2- 0x- 25= (5x^2- 5x^2)- (8x- 0x)- (13- 25)= 0x^2- 8x+ 12[/tex]. That is, [tex]x^2- 8x- 13[/tex] divided by [tex]x^2- 5[/tex] has quotient 5 with remainder [tex]-8x+ 12[/tex]. That remainder is still divided by [tex]x^2- 5[/tex] so can be written as the fraction [tex]\frac{12- 8x}{x^2- 5}[/tex].


So [tex]\frac{5x^2- 8x- 13}{x^2- 5}= 5+ \frac{12- 8x}{x^2- 5}[/tex].

The limit has nothing at all to do with that!

[HallsofIvy]

Well, strictly speaking, you don't go from one to the other because you have a "lim" in one that you don't have in the other! But you go from [tex]\frac{5x^2- 8x- 13}{x^2- 5}[/tex] to [tex]5+ \frac{12- 8x}{x^2- 5}[/tex] by division! A numerical example is that 17 divides into 385 22 times with remainder 12: [tex]\frac{385}{17}= 22+ \frac{12}{17}[/tex].

The numerator has [tex]5x^2[/tex] while the denominator has only [tex]x^2[/tex]. The "trial divisor" is [tex]\frac{5x^2}{x^2}= 5[/tex]. Then you subtract the entire divisor, [tex]x^2- 5[/tex],times 5, from the dividend, [tex]5x^2- 8x- 13[/tex]:

[tex]5x^2- 8x- 13- 5(x^2+ 0x- 5)= 5x^2- 8x- 13- 5x^2- 0x- 25= (5x^2- 5x^2)- (8x- 0x)- (13- 25)= 0x^2- 8x+ 12[/tex]. That is, [tex]x^2- 8x- 13[/tex] divided by [tex]x^2- 5[/tex] has quotient 5 with remainder [tex]-8x+ 12[/tex]. That remainder is still divided by [tex]x^2- 5[/tex] so can be written as the fraction [tex]\frac{12- 8x}{x^2- 5}[/tex].


So [tex]\frac{5x^2- 8x- 13}{x^2- 5}= 5+ \frac{12- 8x}{x^2- 5}[/tex].

The limit has nothing at all to do with that!