You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an alternative browser.
You should upgrade or use an alternative browser.
Limits
- Thread starter calhelp11
- Start date
BigGlenntheHeavy
Senior Member
- Joined
- Mar 8, 2009
- Messages
- 1,577
\(\displaystyle One \ way, \ are \ you \ familiar \ with \ the \ Marqui?\)
BigGlenntheHeavy
Senior Member
- Joined
- Mar 8, 2009
- Messages
- 1,577
\(\displaystyle What!! \ You \ never \ head \ of \ the \ Marqui \ Guillaume \ Francois \ Antoine \ De \ L'Hopital?\)
\(\displaystyle Verily, \ you \ jest.\)
\(\displaystyle Verily, \ you \ jest.\)
BigGlenntheHeavy
Senior Member
- Joined
- Mar 8, 2009
- Messages
- 1,577
\(\displaystyle \lim_{x\to1}\frac{5x^{2}-7x+2}{x^{2}-1} \ gives \ the \ indeterminate \ form \ \frac{0}{0}.\)
\(\displaystyle Hence, \ applying \ "L'Hopital \ rule" \ in \ which \ we \ take \ the \ derivatives \ of \ the \ terms \ separately,\)
\(\displaystyle we \ get \ D_x[5x^{2}-7x+2] \ = \ 10x-7 \ and \ D_x[x^{2}-1] \ = \ 2x.\)
\(\displaystyle Now, \ \lim_{x\to1}\frac{10x-7}{2x} \ = \ \lim_{x\to1}\frac{10-7}{2} \ = \ \frac{3}{2}.\)
\(\displaystyle Hence, \ applying \ "L'Hopital \ rule" \ in \ which \ we \ take \ the \ derivatives \ of \ the \ terms \ separately,\)
\(\displaystyle we \ get \ D_x[5x^{2}-7x+2] \ = \ 10x-7 \ and \ D_x[x^{2}-1] \ = \ 2x.\)
\(\displaystyle Now, \ \lim_{x\to1}\frac{10x-7}{2x} \ = \ \lim_{x\to1}\frac{10-7}{2} \ = \ \frac{3}{2}.\)
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.
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.