Prove that
![\[ \lim_{x \to 0} (1+x)^c = 1 + cx + o(x). \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-d808f1d5d5b450af697d593d235e1ac9_l3.png "Rendered by QuickLaTeX.com")
Using this fact compute
![\[ \lim_{x \to +\infty} \left( \sqrt{x^4 + x^2} - x^2 \right). \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-9390996afa1751c9509bd4a1a7c228ca_l3.png "Rendered by QuickLaTeX.com")
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*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-0d329204f13599ade959ecc6522648a0_l3.png "Rendered by QuickLaTeX.com")
Then, we know (Examples on page 287 of Apostol, taking 
) that
![\[ \lim_{x \to 0} \log(1+x) = x - o(x). \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-0b949e0e8cbbd282d678701dbee84880_l3.png "Rendered by QuickLaTeX.com")
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*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-a126f3ff8e38ce6f11df270d24055835_l3.png "Rendered by QuickLaTeX.com")
Finally we use the expansion, 
, and Theorem 7.8 (page 288 of Apostol on the algebra of the 
-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*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-7593b396fb49fb877f5b977ed898d98d_l3.png "Rendered by QuickLaTeX.com")
Now, to use this to evaluate the requested limit we make the substitution 
. Then 
as 
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*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-3c5b47d6a9c05577250f19761e60abc2_l3.png "Rendered by QuickLaTeX.com")
For part a, could you do a first degree taylor series on (1+x)^c (only taking the first derivative and grouping everything else into o(x), or would that violate some theorem/assumption since we’re it to a polynomial instead of a non-polynomial function?