# need help plz limits

• Dec 7th 2006, 05:23 AM
gracy
need help plz limits
need help
• Dec 7th 2006, 06:03 AM
Soroban
Hello, Gracy!

Are we allowed to use L'Hopital's Rule?

Quote:

7. Evaluate: .$\displaystyle \lim_{x\to0}\frac{e^x - 1}{x}$

. . $\displaystyle (a)\;\infty\qquad(b)\;0\qquad(c)\;1\qquad(d)\;2\qq uad(e)\text{ limit does not exist}$

The function goes to $\displaystyle \frac{0}{0}$ . . . We can use L'Hopital.

L'Hopital: .$\displaystyle \lim_{x\to0}\frac{e^x}{1}\;=\;\frac{e^0}{1}\;=\;1$ . . . answer (c)

Quote:

8. Evaluate: .$\displaystyle \lim_{x\to\infty}\frac{\ln(x^2)}{x^{\frac{1}{2}}}$

. . $\displaystyle (a)\;\infty\qquad(b)\;0\qquad(c)\;2\qquad(3)\text{ limit does not exist}$

The function goes to $\displaystyle \frac{\infty}{\infty}$ . . . We can use L'Hopital.

We have: .$\displaystyle \lim_{x\to\infty}\frac{2\ln(x)}{x^{\frac{1}{2}}}$

L'Hopital: .$\displaystyle \lim_{x\to\infty}\frac{2\cdot\frac{1}{x}}{\frac{1} {2}x^{-\frac{1}{2}}} \;=\;\lim_{x\to\infty}\frac{4}{x^{\frac{1}{2}}} \:=\:0$ . . . answer (b)