L'Hopital's Rule

• May 1st 2010, 06:17 AM
acevipa
L'Hopital's Rule
How would I use L'Hopital's Rule to calculate the following limits. I know the rule and can apply it, it's just that, I'm having trouble with some of these questions.

$\displaystyle 1)\ \lim_ {x \rightarrow 0^+}\ x^\frac{2}{\ln x}$

$\displaystyle 2)\ \lim_{x \rightarrow \infty}\ x^ \frac{1}{x}$

$\displaystyle 2)\ \lim_{x \rightarrow \infty}\ a^ \frac{1}{x},\ a>0$
• May 1st 2010, 06:44 AM
Prove It
Quote:

Originally Posted by acevipa
How would I use L'Hopital's Rule to calculate the following limits. I know the rule and can apply it, it's just that, I'm having trouble with some of these questions.

$\displaystyle 1)\ \lim_ {x \rightarrow 0^+}\ x^\frac{2}{\ln x}$

$\displaystyle 2)\ \lim_{x \rightarrow \infty}\ x^ \frac{1}{x}$

$\displaystyle 2)\ \lim_{x \rightarrow \infty}\ a^ \frac{1}{x},\ a>0$

For these you will need to use the exponential-logarithmic transformation.

1) $\displaystyle x^{\frac{2}{\ln{x}}}= e^{\ln{\left(x^{\frac{2}{\ln{x}}}\right)}}$

$\displaystyle = e^{\left(\frac{2}{\ln{x}}\right)\ln{x}}$

$\displaystyle = e^2$.

Therefore $\displaystyle \lim_{x \to 0^{+}}x^{\frac{2}{\ln{x}}} = \lim_{x \to 0^{+}}e^2$

$\displaystyle = e^2$.

2) $\displaystyle x^{\frac{1}{x}} = e^{\ln{\left(x^{\frac{1}{x}}\right)}}$

$\displaystyle = e^{\left(\frac{1}{x}\right)\ln{x}}$

$\displaystyle = e^{\frac{\ln{x}}{x}}$.

Therefore $\displaystyle \lim_{x \to \infty}x^{\frac{1}{x}} = \lim_{x \to \infty}e^{\frac{\ln{x}}{x}}$

$\displaystyle = e^{\lim_{x \to \infty}\frac{\ln{x}}{x}}$

$\displaystyle = e^{\lim_{x \to \infty} \frac{\left(\frac{1}{x}\right)}{1}}$ by L'Hospital's Rule

$\displaystyle = e^{\lim_{x \to \infty}\frac{1}{x}}$

$\displaystyle = e^0$

$\displaystyle = 1$.

3) Surely $\displaystyle \lim_{x \to \infty}a^{\frac{1}{x}} = a^0 = 1$...