I dont understand why I am getting 2 and the textbook says it is -2.
$$\lim_{x\to 0} \frac{1-e^x}{\sqrt{1+x}-1}$$
I subbed the power series for $e^x$ and $(1+x)^{1/2}$ then got rid of the $1$ on top and on the bottom by taking the $P_0(x)$ of both power series.
$$\lim_{x\to 0} \frac{\sum_{k=1}^{\infty}\frac{x^k}{k!}}{\sum_{k=1}^{\infty}\binom{1/2}{k}x^k}$$
Then took $P_1(x)$ out from the top and bottom. I think my mistake is somewhere here. I dont understand if I can take $P_2(x)$ out of the series like that and modify the series.
$$\lim_{x\to 0} \frac{x + \frac{x^2}{2}\sum_{k=2}^{\infty}\frac{x^{k-2}}{k!}2}{\frac{1}{2}x-\frac{x^2}{8}\sum_{k=2}^{\infty}\binom{1/2}{k}x^{k-2}}$$
So then I divided everything by $x$ and subbed $0$ for the remaining $x$'s which led me to $\frac{1}{2}$
Best Answer
When $x \gt 0$ the numerator is negative and the denominator is positive. When $x \lt 0$ the numerator is positive and the denominator is negative. These mean the limit must be less than zero. In going from the first equation to the second you lost the negative sign on the $e^x$.