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L'Hopital's Rule, 2 Confusing ones

  1. Feb 15, 2006 #1

    So I'm having a bit of difficulty with two of these L'Hospital's Rule problems...

    The first:
    [tex]\mathop {\lim }\limits_{x \to \infty} (\sqrt{x^2 + x} - x)[/tex]

    So when you have an indefinite form [tex]\infty - \infty[/tex], you've got to turn it into a product indefinite form, usually something like [tex]\infty * 0[/tex]

    So I take that,
    [tex]\mathop {\lim }\limits_{x \to \infty} (\sqrt{x^2 + x} - x) = \mathop {\lim }\limits_{x \to \infty} x(\frac{\sqrt{x^2 +x}}{x} - 1)[/tex]

    From there, I try to take the limit of the fraction in the parenthesis, but end up going in circles with it. After I differentiate it the first time and get another indefinite form, I differentiate again and from there, it loops. I keep getting the reciprocal of what I started with. I know that the answer is supposed to come out to be 1/2, but I have no idea how to get there. Any ideas?

    The second one I'm having difficulty with is,
    [tex]\mathop {\lim }\limits_{x \to 1+} lnx tan(\pi x/2)[/tex]

    I'm getting a similar problem with this one. No matter which way I go when I try to turn this one into a fraction, I either get the wrong answer (-1), or it just gets continually more complex. ...The answer here is supposed to be [tex]-2/\pi[/tex]

    Any pointers?
  2. jcsd
  3. Feb 16, 2006 #2


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    And after that, you gotta turn it into [itex]\infty / \infty [/itex] or [itex]0 / 0 [/itex], because that's when l'Hospital applies.

    Useful hint:

    [tex]x = \frac{1}{\frac{1}{x}}[/tex]
  4. Feb 16, 2006 #3


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    Again, you get the indeterminate form [itex]0 \cdot \infty[/itex] but you cannot apply L'Hospital then. As quasar987 pointed out, you need something of the form 0/0 or ∞/∞: use his hint to obtain such a form.
  5. Feb 16, 2006 #4


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    Useful hint

    [tex] \lim_{x\rightarrow +\infty} \left(\sqrt{x^{2}+x}-x\right)=\lim_{x\rightarrow +\infty} \left(\left(\sqrt{x^{2}+x}-x\right) \cdot \frac{\sqrt{x^{2}+x}+x}{\sqrt{x^{2}+x}+x}\right) [/tex]

  6. Feb 16, 2006 #5


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    For the second

    [tex] \lim_{x\rightarrow 1^{+}}\ln x\cdot \tan\frac{\pi x}{2} [/tex]

    can be solved easily if you do these steps

    1. Make the substitution [itex] x=1+y [/itex]. It follows that the new limit will be

    [tex] \lim_{y\rightarrow 0^{+}} \left(\mbox{something}\right) [/tex]

    2. Use the approximation (coming from the MacLaurin expansion of the natural logarithm)

    [tex] \ln\left(1+y)\simeq y [/tex], valid when "y" is very close to 0.

    3.Use the definition of tangent wrt to sine & cosine.

    4.Use the fact that

    [tex] \lim_{y\rightarrow 0^{+}} \sin {}\frac{\pi}{2}\left(y+1\right) = 1 [/tex].

    5. Use l'Hôpital's rule.

    Final result, of course [itex]-\frac{2}{\pi} [/itex].

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