$\displaystyle \lim_{x \to 0} \frac{\ln(1+\sin(2x^5))}{\tan^3(3x) \cdot (5^{x^{2}} - 1)}$ without using L'Hospital or Taylor's series Find limit without using L'Hospital or Taylor's series:
$$\displaystyle \lim_{x \to 0} \frac{\ln(1+\sin(2x^5))}{\tan^3(3x) \cdot (5^{x^{2}} - 1)}$$

$$\displaystyle \lim_{x \to 0} \frac{\ln(1+\sin(2x^5))}{\tan^3(3x) \cdot (5^{x^{2}} - 1)} = \displaystyle \lim_{x \to 0} \frac{\ln(1+\sin(2x^5))}{\sin(2x^5)} \cdot \frac{\sin(2x^5)}{\tan^3(3x) \cdot (5^{x^{2}} - 1)} = $$
$$= \displaystyle \lim_{x \to 0} \frac{\ln(1+\sin(2x^5))}{\sin(2x^5)} \cdot \frac{\sin(2x^5)}{x^5} \cdot \frac{x^2}{5^{x^{2}} - 1} \cdot \frac{ x^3}{\tan^3(3x)}$$

*

*$\displaystyle \lim_{x \to 0} \frac{\ln(1+\sin(2x^5))}{\sin(2x^5)} = 1$

*$\displaystyle \lim_{x \to 0} \frac{\sin(2x^5)}{x^5} = 2$

*$\displaystyle \lim_{x \to 0} \frac{x^2}{5^{x^{2}} - 1} = \frac{1}{\ln5}$

*$\displaystyle \lim_{x \to 0} \frac{ x^3}{\tan^3(3x)} = \displaystyle \lim_{x \to 0} \cos^3(3x) \cdot \displaystyle \lim_{x \to 0} \frac{x^3}{\sin^3(3x)} = 1 \cdot \left( \frac{1}{3} \right)^3 = \frac{1}{27}$
We get:
$$\displaystyle \lim_{x \to 0} \frac{\ln(1+\sin(2x^5))}{\tan^3(3x) \cdot (5^{x^{2}} - 1)} = 1 * 2 * \frac{1}{\ln5} * \frac{1}{27} = \frac{2}{27 \ln5}$$
All credits to @Bernard. Thank you all, question closed.

Just a slight variant on Bernard's answer: rewrite the fraction as
$${\ln(1+\sin(2x^5))\over\sin(2x^5)}\cdot{\sin(2x^5)\over2x^5}\cdot\left(3x\over\sin(3x)\right)^3\cdot{x^2\ln5\over e^{x^2\ln5}-1}\cdot{2\cos^3(3x)\over3^3\ln5}$$
then use the familiar limits
$$\lim_{u\to0}{\ln(1+u)\over u}=\lim_{u\to0}{\sin u\over u}=\lim_{u\to0}{e^u-1\over u}=1$$
to get the limit
$$1\cdot1\cdot1^3\cdot1\cdot{2\cdot1^3\over27\ln5}={2\over27\ln5}$$
