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Find values for a constant such that a given limit will exist

Consider the limit expression

    \[ \lim_{x \to 0} \frac{e^{ax} - e^x - x}{x^2}. \]

Find the value of the constant a such that this limit will be finite. Find the value of the limit in this case.

We have

    \begin{align*}  \lim_{x \to 0} \frac{e^{ax} - e^x - x}{x^2} &= \lim_{x \to 0} \left( \frac{1}{x^2} \left( 1 + ax + \frac{a^2x^2}{2} + o(x^2) - 1 - x - \frac{x^2}{2} + o(x^2) - x \right) \right)\\[9pt]  &= \lim_{x \to 0} \left( \frac{1}{x^2} \left( (a-2)x + \frac{(a^2-1)x^2}{2} + o(x^2) \right) \right) \\[9pt]  &= \lim_{x \to 0} \left( \frac{a-2}{x} + \frac{a^2-1}{2} + o(1) \right). \end{align*}

In order for this limit to exist we must have a-2 = 0; hence a = 2. The limit is then

    \[ \lim_{x \to 0} \left( \frac{a^2-1}{2} + o(1) \right) = \frac{3}{2} + 0 = \frac{3}{2}. \]


  1. tom says:

    The answer doesn’t seem right, although it’s the same as the book answer. I’m having a problem seeing how o(1)=1. o(1) means the function will go to zero for any value denominator e>0, doesn’t it? If o(1) did mean 1, then couldn’t we apply any arbitrary constant we want?

    • RoRi says:

      The answer is right, but I made a mistake in the way I derived (and then an offsetting mistake that got me back to the right answer). As you pointed out \lim_{x \to 0} o(1) = 0. But, in a previous line I dropped a square on the a for one of the coefficients. We had \frac{(ax)^2}{2} in the expansion of e^{ax}, but then when I combined the coefficients for x^2 I said it was \frac{(a-1)}{2} instead of \frac{a^2-1}{2}. When we take them limit we then get the desired \frac{3}{2}. I think it’s all fixed now. Let me know if there’s still a mistake.

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