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Prove that (1+x)c = 1 + cx + o(x) as x goes to 0

Prove that

    \[ \lim_{x \to 0} (1+x)^c = 1 + cx + o(x). \]

Using this fact compute

    \[ \lim_{x \to +\infty} \left( \sqrt{x^4 + x^2} - x^2 \right). \]


Proof. We use the definition and continuity of the exponential to evaluate the limit,

    \begin{align*}  \lim_{x \to 0} (1+x)^c &= \lim_{x \to 0} e^{ c \log (1+x)} \\[9pt]  &= \exp \left( \lim_{x \to 0} c \log (1+x) \right). \end{align*}

Then, we know (Examples on page 287 of Apostol, taking n = 1) that

    \[ \lim_{x \to 0} \log(1+x) = x - o(x). \]

Therefore,

    \begin{align*}  \lim_{x \to 0} (1+x)^c &= \exp \left(\lim_{x \to 0} c \log(1+x) \right) \\[9pt]  &= e^{ cx - o(x) }. \end{align*}

Finally we use the expansion, e^x = 1 + x + o(x), and Theorem 7.8 (page 288 of Apostol on the algebra of the o-symbols) to conclude,

    \begin{align*}  \lim_{x \to 0}(1+x)^c &= e^{cx - o(x)} \\[9pt]  &= 1 + (cx - o(x)) + o(cx-o(x)) \\[9pt]  &= 1 + cx + o(x). \qquad \blacksquare \end{align*}

Now, to use this to evaluate the requested limit we make the substitution t = \frac{1}{x^2}. Then t \to 0 as x \to +\infty and we have

    \begin{align*}  \lim_{x \to +\infty} \left( \sqrt{x^4+x^2} - x^2 \right) &= \lim_{t \to 0} \left( \sqrt{ \frac{1}{t^2} + \frac{1}{t} } - \frac{1}{t} \right) \\[9pt]  &= \lim_{t \to 0} \left( \frac{1}{t} \sqrt{1 + t} - \frac{1}{t} \right) \\[9pt]  &= \lim_{t \to 0} \frac{(1+t)^{\frac{1}{2}} - 1}{t} \\[9pt]  &= \lim_{t \to 0} \frac{1 + \frac{1}{2} t + o(t) - 1}{t} \\[9pt]  &= \lim_{t \to 0} \left( \frac{1}{2} + \frac{o(t)}{t} \right) \\[9pt]  &= \frac{1}{2}. \end{align*}

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