Huge hint: \begin{equation} (\sqrt{x}-1)(\sqrt{x}+1) = x-1 \end{equation}

Another way to think of it is to break up the fraction as follows: \begin{align} \frac{\sqrt{x} + 1 }{x-1} = (\sqrt{x}+1) \frac{1}{x-1} \end{align} from which we can see that: As x tends to 1 ...

\frac{\sqrt{x + 1}}{x} =\frac{x + 1}{x\sqrt{x + 1}} =\frac{1}{\sqrt{x + 1}} + \frac{1}{x\sqrt{x + 1}} \to -\infty \quad \text{as} \quad x \to 0^-

I repost it because I encountered a fatal error on the first step \left|\frac{\sqrt{x+5}-2}{x+1}-\frac{1}{4}\right| =\left|\frac{4\sqrt{x+5}-x-9}{4(x+1)}\right| =\left|\frac{16(x+5)-(x+9)^2}{4(x+1)(\sqrt{x+5}+x+9)}\right| ...

The limit is the expression evaluated at 3. Explanation: \displaystyle\lim_{{{x}\to{3}}}\frac{\sqrt{{{x}+{1}}}}{{{x}-{4}}}=\frac{\sqrt{{{3}+{1}}}}{{{3}-{4}}} \displaystyle\lim_{{{x}\to{3}}}\frac{\sqrt{{{x}+{1}}}}{{{x}-{4}}}=\frac{\sqrt{{{4}}}}{{-{{1}}}} ...

Revised to avoid l’Hospital’s rule: Your second one can be finished off like this: \begin{align*} \lim_{x\to 0}\frac{-2\sin 2x\sin x}{x^2}&=-2\left(\lim_{x\to 0}\frac{\sin 2x}x\right)\left(\lim_{x\to 0}\frac{\sin x}x\right)\\ &=-4\left(\lim_{x\to 0}\frac{\sin 2x}{2x}\right)\cdot1\\ &=-4\;. \end{align*} ...