Strange mistakes when calculate limits I have difficulties with calculating the following limits. W|A gives the correct answers for both of them:
$$
\lim_{x \to +\infty} \sqrt{x} \cdot \left(\sqrt{x+\sqrt x} + \sqrt{x - \sqrt x} - 2\sqrt x\right) = \lim_{x \to +\infty} \sqrt{x^2+x\sqrt x} + \sqrt{x^2-x\sqrt x} - 2x = \\
= \lim_{x \to +\infty} x\left(\sqrt{1+ \frac {\sqrt x}{x}} + \sqrt{1 - \frac{\sqrt x}{x}} - 2\right) = \left[t = \frac 1x\right] = \lim_{t \to 0+} \frac{\sqrt{1+\sqrt t} - 1 +\sqrt{1-\sqrt t} - 1}{t} = 
 \left[c = \sqrt t\right] = \lim_{c \to 0+} \frac{\sqrt{1+c} - 1 +\sqrt{1-c} - 1}{c^2} = \frac 12c + \left(-\frac 12 c\right) = 0
$$
It's because:
$$
\lim_{c \to 0+} \frac{\sqrt{1+c}-1}{c} = \left|k = \sqrt{1+c}\right| = \lim_{k \to 1+} \frac{k-1}{k^2-1} = \lim_{k \to 1+} \frac{1}{k+1} = \frac 12
$$
The second part is similar to it. But the answer is -$\frac 14$.
And the second limit is:
$$
\lim_{x \to 0} \left(\frac{e^x - \sin x}{\sqrt{1 - 2x} + \log(x+1)}\right)^\cfrac{1}{x^2} = \exp\left(\lim_{x \to 0} \frac{1}{x^2}\left(\frac{e^x-\sin x}{\sqrt{1-2x} + \log(x+1)}-1\right)\right) = 
 \exp\left(\lim_{x \to 0}\frac{e^x - \sin x - \sqrt{1-2x} - \log(x+1)}{x^2} \cdot \frac{1}{\sqrt{1-2x}+\log(x+1)}\right) = \exp\left(\lim_{x \to 0} \left(\frac{e^x-1}{x \cdot x} - \frac{\sin x}{x \cdot x} - \frac{\sqrt{1-2x}-1}{x \cdot x} - \frac{\log(x+1)}{x \cdot x}\right)\right) = \exp\left(\lim_{x \to 0} \left(\frac 1x - \frac 1x + \frac 1x - \frac 1x\right)\right) = e^0 = 1
$$
The answer is $e^\frac 32$. Where am I wrong?
 A: In the first line of the first one: 
$\sqrt{x}\sqrt{x + \sqrt{x}} = \sqrt{x^2 + x\sqrt{x}}$, not $\sqrt{x^2 + x}$.
A: How do you get from $$\lim\limits_{c \rightarrow 0^+} \frac{\sqrt{1+c} - 1 + \sqrt{1-c} - 1}{c^2}$$ to $$\frac{1}{2}c - \frac{1}{2}c?$$

You cannot just evaluate part of the limit and leave another part. Compare your mistake with "evaluating" $\lim_{x \rightarrow 0} \frac{x^2}{x^2}$ by noting that $\lim_{x \rightarrow 0} \frac{x^2}{x}$ is zero and then concluding that $$\lim_{x \rightarrow 0} \frac{x^2}{x^2} = \lim_{x \rightarrow 0} \frac{0}{x} = 0$$

In terms of Taylor series, $\frac{\sqrt{1+x} - 1}{x}$ is approximated by $\frac{1}{2} - \frac{x}{8} + O(x^2)$. So the limit as $x \rightarrow 0$ is equal to $\frac{1}{2}$, but  $\frac{\sqrt{1+x} - 1}{x^2}$ is approximated by the Taylor series $\frac{1}{2x} - \frac{1}{8} + O(x)$. Notice that this is not equal to $\frac{1}{2x}$, which you substitute it with! In a similar way, $\frac{\sqrt{1-x} - 1}{x}$ is approximated by the Taylor series $-\frac{1}{2x} - \frac{1}{8} + O(x)$, and adding these up gives $-\frac{1}{4} + O(x)$, and thus the limit of $\frac{\sqrt{1+x} - 1 + \sqrt{1 - x} - 1}{x^2}$ as $x \rightarrow 0$ is equal to -$\frac{1}{4}$.
A: For the first problem, divide everything by x first. You have now terms which look like Sqrt[1+a] where "a" is small. Apply the Taylor series, that is to say that, close to a=0, Sqrt[1+a] can be approximated by (1 + a / 2 - a^2 / 8). Are you able to continue with this ?
