Need of L-Hospital's Rule

Calculus Level pending

With the following flawed work out I strived to prove that

lim ⁡ x → 0 x x = ∞ \lim_{x \to 0} x^x= \infty

Which of the following step is made firstly incorrect?

Let a function f ( x ) = x x f(x) = x^x , where x > 0 x>0 .

Step :

  1. lim ⁡ x → 0 + f ( x ) \displaystyle \lim_{x \to 0^+} f(x)
  2. lim ⁡ x → 0 + x x \displaystyle \lim_{x \to 0^+} x^x
  3. lim ⁡ x → 0 + e log ⁡ x x \displaystyle \lim_{x \to 0^+} e^{\log x^x}
  4. lim ⁡ x → 0 + e x log ⁡ x \displaystyle \lim_{x \to 0^+} e^{x\log x}
  5. e lim ⁡ x → 0 + ( x log ⁡ x ) e^{\lim_{x \to 0^+}(x \log x)}
  6. e lim ⁡ x → 0 + d d x ( x log ⁡ x ) e^{\lim_{x \to 0^+} \frac d{dx}(x \log x)}
  7. e lim ⁡ x → 0 + ( 1 + log ⁡ x ) e^{\lim_{x \to 0^+} (1+\log x)}
  8. e ∞ e^\infty
  9. ∞ \infty (Hence proved)
7 5 2 8 4 6 1

Naren Bhandari by Naren Bhandari , 21, India

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1 solution

If we apply L'Hospital, we need to differentiate the denominator as well!

So Step 6 is wrong!

lim ⁡ x → 0 + ( x log ⁡ x ) = lim ⁡ x → 0 + log ⁡ x 1 / x = lim ⁡ x → 0 + d d x log ⁡ x d d x 1 x \lim_{x \to 0^+} (x \log x) = \lim_{x \to 0^+} \frac{ \log x}{1/x} =\lim_{x \to 0^+} \frac {\frac{d}{dx} \log x} { \frac {d} {dx} \frac{1} {x}}

This is how we will apply L'Hospital

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It is not clear to me how L'hopital is involved.

Instead, the error is because of (to be filled in).

Calvin Lin Staff - 4 years, 3 months ago

If we differentiate the denominator as you mentioned which carries constant 1 1 results to ∞ ∞ so better is to mentioned that we need to get that form so that L-hopital rule is involved as e lim ⁡ x → 0 ( l o g x 1 / x e^{\lim_{x→0}(\frac{logx}{1/x}} .

Naren Bhandari - 4 years, 3 months ago

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Ah ic. Could you make that usage of L-hopital's rule much clearer?

Otherwise, it seems to me that you're saying lim ⁡ f ( x ) = lim ⁡ f ′ ( x ) \lim f(x) = \lim f'(x) , where f ( x ) = x log ⁡ x f(x) = x \log x (and that's the error that I thought was happening).

Calvin Lin Staff - 4 years, 3 months ago

  1. lim ⁡ x → 0 + f ( x ) \displaystyle \lim_{x \to 0^+} f(x)
  2. lim ⁡ x → 0 + x x \displaystyle \lim_{x \to 0^+} x^x
  3. lim ⁡ x → 0 + e log ⁡ x x \displaystyle \lim_{x \to 0^+} e^{\log x^x}
  4. lim ⁡ x → 0 + e x log ⁡ x \displaystyle \lim_{x \to 0^+} e^{x\log x}
  5. e lim ⁡ x → 0 + ( x log ⁡ x ) e^{\lim_{x \to 0^+}(x \log x)}
  6. e^{\lim_{x \to 0^+} (\frac{logx}{1/x})
Now we can make the usage of L-hopital rule as the form is ∞ ∞ \frac{∞}{∞}

  1. e lim ⁡ x → 0 + x − 1 − x − 2 e^{\lim_{x \to 0^+} \frac {x^{-1}}{-x^{-2}}}

  2. e l i m x → 0 + ( − x ) e^{lim_{x \to 0^+} (-x) }

  3. e 0 = 1 e^0 = 1

Naren Bhandari - 4 years, 3 months ago

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