L'Hopital's Rule: lim[x -> 1] [(x e^x - e)/(x^2 - 1)]: how did they get the exponent?

[Teller22]

L'Hopital's Rule: lim[x -> 1] [(x e^x - e)/(x^2 - 1)]: how did they get the exponent?

Hi, I was working on this problem on Khan Academy and am wondering how they got the answer they did.

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. . .\(\displaystyle \displaystyle \lim_{x \rightarrow 1}\, \dfrac{x\, e^x\, -\, e}{x^2\, -\, 1}\)

. . . . . .\(\displaystyle \displaystyle =\, \lim_{x \rightarrow 1}\, \dfrac{\dfrac{d}{dx}\left[x\, e^x\, -\, e\right]}{\dfrac{d}{dx}\left[x^2\, -\, 1\right]}\qquad \mbox{ L'Hopital's rule}\)

. . . . . .\(\displaystyle \displaystyle =\,\lim_{x \rightarrow 1}\, \dfrac{e^x\, +\, x\, e^x}{2x}\)

. . . . . .\(\displaystyle \displaystyle =\, \dfrac{e^{(1)}\, +\, (1)\, e^{(1)}}{2(1)}\qquad \mbox{ Substitution}\)

. . . . . .\(\displaystyle \displaystyle =\, e\)

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My problem with their solution is how the negative e suddenly gained an x exponent (I know e stays the same after a derivative so is it just symbolic?) and also how it became positive.

Any help would be greatly appreciated.

 

[tkhunny]

Hi, I was working on this problem on Khan Academy and am wondering how they got the answer they did.

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. . .\(\displaystyle \displaystyle \lim_{x \rightarrow 1}\, \dfrac{x\, e^x\, -\, e}{x^2\, -\, 1}\)

. . . . . .\(\displaystyle \displaystyle =\, \lim_{x \rightarrow 1}\, \dfrac{\dfrac{d}{dx}\left[x\, e^x\, -\, e\right]}{\dfrac{d}{dx}\left[x^2\, -\, 1\right]}\qquad \mbox{ L'Hopital's rule}\)

. . . . . .\(\displaystyle \displaystyle =\,\lim_{x \rightarrow 1}\, \dfrac{e^x\, +\, x\, e^x}{2x}\)

. . . . . .\(\displaystyle \displaystyle =\, \dfrac{e^{(1)}\, +\, (1)\, e^{(1)}}{2(1)}\qquad \mbox{ Substitution}\)

. . . . . .\(\displaystyle \displaystyle =\, e\)

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My problem with their solution is how the negative e suddenly gained an x exponent (I know e stays the same after a derivative so is it just symbolic?) and also how it became positive.

Any help would be greatly appreciated.

e did not suddently gain an exponent. \(\displaystyle \dfrac{d}{dx}e = 0\)

Your mysterious 'e' actually just vanished. Now you have to figure out where the \(\displaystyle e^{x}\) came from. Think "Product Rule".

Note: Substitution is bad.

 

[Dr.Peterson]

Hi, I was working on this problem on Khan Academy and am wondering how they got the answer they did.

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. . .\(\displaystyle \displaystyle \lim_{x \rightarrow 1}\, \dfrac{x\, e^x\, -\, e}{x^2\, -\, 1}\)

. . . . . .\(\displaystyle \displaystyle =\, \lim_{x \rightarrow 1}\, \dfrac{\dfrac{d}{dx}\left[x\, e^x\, -\, e\right]}{\dfrac{d}{dx}\left[x^2\, -\, 1\right]}\qquad \mbox{ L'Hopital's rule}\)

. . . . . .\(\displaystyle \displaystyle =\,\lim_{x \rightarrow 1}\, \dfrac{e^x\, +\, x\, e^x}{2x}\)

. . . . . .\(\displaystyle \displaystyle =\, \dfrac{e^{(1)}\, +\, (1)\, e^{(1)}}{2(1)}\qquad \mbox{ Substitution}\)

. . . . . .\(\displaystyle \displaystyle =\, e\)

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My problem with their solution is how the negative e suddenly gained an x exponent (I know e stays the same after a derivative so is it just symbolic?) and also how it became positive.

Any help would be greatly appreciated.

I would guess that you didn't actually try doing the differentiation yourself, which is why you imagined that it was the "- e" term from which the second term of the derivative came. I hope you can see now that the two terms both arise from the product rule applied to the first term of the numerator; the derivative of the second term, which is constant, is zero.

Your "substitution" is valid, given that you have first observed that the new function is continuous at x=1 (unlike the original function).