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Example question using Taylor series with Limits

How do you go about solving limits like the one shown above? Why is it not correct to express the taylor series in terms of y instead of x, and then multiply the taylor series by $\sqrt{y}$, followed by subtracting y? Like so:

$\sqrt{y^2+y} = (1+\frac{1}{2}y-\frac{1}{8}y^2)\sqrt{y} - y$

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  • $\begingroup$ Your expansion is not valid as binomial theorem would require $|y|<1$. $\endgroup$ – Paramanand Singh Apr 5 at 4:05
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When $y$ tends to $+\infty$, $$\sqrt{y^2+y} - y = y \left( \sqrt{1+\frac{1}{y}}-1\right) = y \left( \frac{1}{2y} + o \left( \frac{1}{y}\right)\right) = \frac{1}{2} + o(1)$$

So the limit is $\frac{1}{2}$.

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