<s>[INST] Evaluating $\lim_{x\to 0} \frac{1-\sqrt{3x+1}}{2-\sqrt{5x+4}}$ $$ \lim_{x\to 0} \frac{1-\sqrt{3x+1}}{2-\sqrt{5x+4}}$$
How do I solve this without using derivatives or integrals.
 [/INST] Both the numerator and the denominator are of the form
$$
a-\sqrt{bx+a^2}
$$
(with $a>0$). In the numerator we have $a=1$, $b=3$; in the denominator we have $a=2$, $b=5$.
Let's try
\begin{align}
\lim_{x\to0}\frac{a-\sqrt{bx+a^2}}{x}
&=\lim_{x\to0}\frac{a-\sqrt{bx+a^2}}{x}
  \frac{a+\sqrt{bx+a^2}}{a+\sqrt{bx+a^2}}\\
&=\lim_{x\to0}\frac{-bx}{x(a+\sqrt{bx+a^2})}\\
&=\lim_{x\to0}\frac{-b}{a+\sqrt{bx+a^2}}\\
&=-\frac{b}{2a}
\end{align}
Thus your limit is
$$
\lim_{x\to 0} \frac{1-\sqrt{3x+1}}{2-\sqrt{5x+4}}=
\lim_{x\to 0} \frac{1-\sqrt{3x+1}}{x}\frac{x}{2-\sqrt{5x+4}}=
\left(-\frac{3}{2}\right)\left(-\frac{4}{5}\right)=\frac{6}{5}
$$

When you'll learn Taylor expansions, you'll see that
$$
\sqrt{1+kx}=1+\frac{1}{2}kx+o(x)
$$
so
$$
\frac{1-\sqrt{1+3x}}{2-\sqrt{4+5x}}=
\frac{1-\sqrt{1+3x}}{2-2\sqrt{1+\frac{5}{4}x}}=
\frac{1-(1+\frac{3}{2}x+o(x))}{2-2(1+\frac{5}{8}x+o(x))}=
\frac{-\frac{3}{2}x+o(x)}{-\frac{5}{4}x+o(x)}=
\frac{-\frac{3}{2}+o(1)}{-\frac{5}{4}+o(1)}
$$
and so the limit is
$$
\frac{-3/2}{-5/4}=\frac{6}{5}
$$
</s>