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Thread: L'Hopital's Rule

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    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.

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

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

    $\displaystyle 2)\ \lim_{x \rightarrow \infty}\ a^ \frac{1}{x},\ a>0$
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  2. #2
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    Quote Originally Posted by acevipa View Post
    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 $\displaystyle \lim_{x \to \infty}a^{\frac{1}{x}} = a^0 = 1$...
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