Example. Compute the limit

$$
\lim _{x \rightarrow 0}\left(\frac{1+x^{2} 2^{x}}{1+x^{2} 5^{x}}\right)^{1 / \sin ^{3} x}
$$
Solution. When $x \rightarrow 0$, the expression under the limit sign represents a power, the base of which tends to one:

$$
\lim _{x \rightarrow 0} \frac{1+x^{2} 2^{x}}{1+x^{2} 5^{x}}=1
$$

and the exponent tends to infinity:

$$
\lim _{x \rightarrow 0} \frac{1}{\sin ^{3} x}=\infty
$$

1. Transform the expression under the limit sign:

$$
\left(\frac{1+x^{2} 2^{x}}{1+x^{2} 5^{x}}\right)^{1 / \sin ^{3} x}=\exp \left(\frac{1}{\sin ^{3} x} \ln \frac{1+x^{2} 2^{x}}{1+x^{2} 5^{x}}\right)
$$

2. Since the exponential function $e^{x}$ is continuous, we can take the limit under the sign of this function. We have

$$
\lim _{x \rightarrow 0}\left(\frac{1+x^{2} 2^{x}}{1+x^{2} 5^{x}}\right)^{1 / \sin ^{3} x}=\exp \left(\lim _{x \rightarrow 0} \frac{1}{\sin ^{3} x} \ln \frac{1+x^{2} 2^{x}}{1+x^{2} 5^{x}}\right)
$$

3. Calculate the limit of the exponent

$$
\lim _{x \rightarrow 0} \frac{1}{\sin ^{3} x} \ln \frac{1+x^{2} 2^{x}}{1+x^{2} 5^{x}}
$$

## Transform the expression under the limit sign to the form

$$
\frac{1}{\sin ^{3} x} \ln \left[1+\frac{x^{2} 5^{x}\left((2 / 5)^{x}-1\right)}{1+x^{2} 5^{x}}\right]
$$

and replace the infinitesimal functions with equivalent ones, we have

$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{1}{\sin ^{3} x} \ln \left[1+\frac{x^{2} 5^{x}\left((2 / 5)^{x}-1\right)}{1+x^{2} 5^{x}}\right]= \\
& \quad=\lim _{x \rightarrow 0} \frac{1}{x^{3}} \frac{x^{2} 5^{x}\left((2 / 5)^{x}-1\right)}{1+x^{2} 5^{x}}=\lim _{x \rightarrow 0} \frac{1}{x^{3}} \frac{x^{2} 5^{x} x \ln (2 / 5)}{1+x^{2} 5^{x}}=\ln \frac{2}{5}
\end{aligned}
$$

4. Finally, we get

$$
\lim _{x \rightarrow 0}\left(\frac{1+x^{2} 2^{x}}{1+x^{2} 5^{x}}\right)^{1 / \sin ^{3} x}=e^{\ln (2 / 5)}=\frac{2}{5}
$$

Answer. $\lim _{x \rightarrow 0}\left(\frac{1+x^{2} 2^{x}}{1+x^{2} 5^{x}}\right)^{1 / \sin ^{3} x}=\frac{2}{5}$.

Conditions of the Problem. Calculate the limit.

1. $\lim _{x \rightarrow 0}\left(\frac{1+x 3^{x}}{1+x 2^{x}}\right)^{1 / x^{2}}$.
2. $\lim _{x \rightarrow 0}\left(\frac{1+x^{2} 3^{x}}{1+x^{2} 4^{x}}\right)^{1 / \operatorname{tg}^{3} x}$.
3. $\lim _{x \rightarrow 0}\left(\frac{1+\sin x \cos 3 x}{1+\sin x \cos 2 x}\right)^{1 / \sin ^{3} x}$.
4. $\lim _{x \rightarrow 0}(1-\ln \cos x)^{1 / \sin ^{2} x}$.
5. $\lim _{x \rightarrow 0}\left(2-5^{\sin ^{2} x}\right)^{1 / x^{2}}$.
6. $\lim _{x \rightarrow 0}(\cos x)^{1 / x^{2}}$.
7. $\lim _{x \rightarrow 0}(\cos x)^{1 /(x \sin x)}$.
8. $\lim _{x \rightarrow 0}\left(1+\sin ^{2} x\right)^{1 / \ln \cos x}$.
9. $\lim _{x \rightarrow 0}\left(\frac{1+x \cos 2 x}{1+x \cos x}\right)^{1 / x^{3}}$.
10. $\lim _{x \rightarrow 0}(2-\cos x)^{1 / \ln \left(1+2 x^{2}\right)}$.

Answers. 1. 3/2. 2. 3/4. 3. $e^{-5 / 2}$. 4. $e^{1 / 2}$. 5. $1 / 5.6 . e^{-1 / 2}$. 7. $e^{-1 / 2}$. 8. $e^{-2}$. 9. $e^{-3 / 2}$. 10. $e^{1 / 4}$.

### 3.11. Calculation of $\lim _{x \rightarrow a}\left[u(x)^{v(x)}\right]$

Problem Statement. Calculate the limit

$$
\lim _{x \rightarrow a}\left[u(x)^{v(x)}\right]
$$

where $\lim _{x \rightarrow a} u(x)=1$ and $\lim _{x \rightarrow a} v(x)=\infty$.

PLAN OF SOLUTION.

1. To use the table of equivalent infinitesimals, make the substitution $t=x-a$ (then $t \rightarrow 0$ as $x \rightarrow a$) and transform the expression under the limit sign:

$$
u(x)^{v(x)}=u_{1}(t)^{v_{1}(t)}=e^{v_{1}(t) \ln u_{1}(t)}
$$

2. Since the exponential function $e^{x}$ is continuous, we can take the limit under the sign of this function. We have

$$
\lim _{x \rightarrow a}\left[u(x)^{v(x)}\right]=e^{\lim m_{t \rightarrow 0}\left[v_{1}(t) \ln u_{1}(t)\right]}
$$

3. Calculate the limit of the exponent

$$
\lim _{t \rightarrow 0}\left[v_{1}(t) \ln u_{1}(t)\right]
$$

replacing the infinitesimal functions with equivalent ones.

4. Write the final answer.