Limits

[calhelp11]

FIND THE LIMIT


Lim ->1

5x^2 -7X +2
___________
x^2 -1



The anwser is 3/2

How did they get this , please explain.

 

[BigGlenntheHeavy]

$\displaystyle One \ way, \ are \ you \ familiar \ with \ the \ Marqui?$

 

[calhelp11]

no

 

[BigGlenntheHeavy]

$\displaystyle What!! \ You \ never \ head \ of \ the \ Marqui \ Guillaume \ Francois \ Antoine \ De \ L'Hopital?$

$\displaystyle Verily, \ you \ jest.$

 

[calhelp11]

If u can help me , it would be much appreciated.... i am really in a time crunch here

 

[BigGlenntheHeavy]

$\displaystyle \lim_{x\to1}\frac{5x^{2}-7x+2}{x^{2}-1} \ gives \ the \ indeterminate \ form \ \frac{0}{0}.$

$\displaystyle Hence, \ applying \ "L'Hopital \ rule" \ in \ which \ we \ take \ the \ derivatives \ of \ the \ terms \ separately,$

$\displaystyle we \ get \ D_x[5x^{2}-7x+2] \ = \ 10x-7 \ and \ D_x[x^{2}-1] \ = \ 2x.$

$\displaystyle Now, \ \lim_{x\to1}\frac{10x-7}{2x} \ = \ \lim_{x\to1}\frac{10-7}{2} \ = \ \frac{3}{2}.$

 

[calhelp11]

thank you !

 

[galactus]

If you have not been exposed to L'Hopital yet:

Factor top and bottom:

$\displaystyle \lim_{x\to 1}\frac{(5x-2)(x-1)}{(x+1)(x-1)}$

Canecl the x-1 and we have:

$\displaystyle \lim_{x\to 1}\frac{5x-2}{x+1}$

Now, plug in x=1 and get your 3/2.

No disrespect intended, but L'Hopital is rather an overkill here. Limits are introduced at the beginning of calc I, but L'Hopital is normally not introduced until calc II. Myself, I only like to use it when confronted with a toughy that is otherwise difficult without it.