53. 54) $\lim _{x \rightarrow 0} \frac{1-\sqrt{x+1}}{x}$;
3) $\lim _{x \rightarrow 0} \frac{\tan x}{1-\sqrt{1+\tan x}}$
4) $\lim _{x \rightarrow 1} \frac{2-\sqrt{x}}{3-\sqrt{2 x+1}}$
5) $\lim _{x \rightarrow 1} \frac{1-\sqrt{x}}{1-\sqrt[3]{x}}$.
Solution. First, we find that the given function represents the ratio of two infinitesimally small quantities (case $\frac{0}{0}$) when the argument is changed. We then transform the fraction to eliminate the factor that tends to zero:

1) We eliminate the irrationality in the numerator by multiplying the numerator and denominator by $1+\sqrt{x+1}$, and then cancel the fraction by $x$:

![](https://cdn.mathpix.com/cropped/2024_05_22_8a069bcd8684ea7ea3dbg-036.jpg?height=94&width=753&top_left_y=1120&top_left_x=265)

$$
\begin{aligned}
& =\lim \frac{-x}{x(1+\sqrt{x+1})}=\lim \frac{-1}{1+\sqrt{x+1}}=-\frac{1}{2} ;
\end{aligned}
$$

2) We multiply the numerator and denominator by the product

$$
(2+\sqrt{x})(3+\sqrt{2 x+1})
$$

and then cancel the fraction by $4-x$:

$$
\lim _{x \rightarrow 4} \frac{2-\sqrt{x}}{3-\sqrt{2 x+1}}=\lim \frac{(4-x)(3+\sqrt{2 x+1})}{(9-2 x-1)(2+\sqrt{x})}=\lim \frac{3+\sqrt{2 x+1}}{2(2+\sqrt{x})}=\frac{3}{4}
$$

3) We multiply the numerator and denominator by $1+\sqrt{1+\operatorname{tg} x}$ and cancel the fraction by $\operatorname{tg} x$:

$$
\begin{gathered}
\lim _{x \rightarrow 0} \frac{\operatorname{tg} x}{1-\sqrt{1+\operatorname{tg} x}}=\lim \frac{\operatorname{tg} x(1+\sqrt{1+\operatorname{tg} x})}{1-1-\operatorname{tg} x}= \\
=-\lim (1+\sqrt{1+\operatorname{tg} x})=-2
\end{gathered}
$$

4) We multiply the numerator and denominator by the product

$$
(1+\sqrt{x})\left(1+\sqrt[3]{x}+\sqrt[3]{x^{2}}\right)
$$

and then cancel the fraction by $1-x$:

$\lim _{x \rightarrow 1} \frac{1-\sqrt{x}}{1-\sqrt[3]{x}}=\lim \frac{(1-x)\left(1+\sqrt[3]{x}+\sqrt[3]{x^{2}}\right)}{(1-x)(1+\sqrt{x})}=\lim \frac{1+\sqrt[3]{x}+\sqrt[3]{x^{2}}}{1+\sqrt{x}}=\frac{3}{2}$.

Alternatively, this problem can be solved by a change of variable. Letting $x=t^{6}$, we get $t \rightarrow 1$ when $x \rightarrow 1$ and

![](https://cdn.mathpix.com/cropped/2024_05_22_8a069bcd8684ea7ea3dbg-037.jpg?height=105&width=1050&top_left_y=416&top_left_x=120)