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

Question

Find limit without using L'Hospital or Taylor's series:

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

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

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

• $\displaystyle \lim_{x \to 0} \frac{\ln(1+\sin(2x^5))}{\sin(2x^5)} = 1$
• $\displaystyle \lim_{x \to 0} \frac{\sin(2x^5)}{x^5} = 2$
• $\displaystyle \lim_{x \to 0} \frac{x^2}{5^{x^{2}} - 1} = \frac{1}{\ln5}$
• $\displaystyle \lim_{x \to 0} \frac{ x^3}{\tan^3(3x)} = \displaystyle \lim_{x \to 0} \cos^3(3x) \cdot \displaystyle \lim_{x \to 0} \frac{x^3}{\sin^3(3x)} = 1 \cdot \left( \frac{1}{3} \right)^3 = \frac{1}{27}$

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.

Answer 1

You can do it using equivalents and some standard limits: rewrite the fraction as $$\frac{\ln(1+\sin(2x^5))}{\sin(2x^5)}\frac{\sin(2x^5)}{\tan^3(3x)} \cdot \frac{x^2}{(5^{x^{2}} - 1)}\cdot\frac1{x^2}.$$ By substitution, the first factor tends to $1$. Also by substitution, $\:\dfrac{5^{x^2}-1}{x^2}$ tends to the derivative of $5^x$ at $0$, i.e. $\ln 5$. Last, $\sin(2x^5)\sim_0 2x^5$ and $\tan 3x)\sim_03x$, so $\tan^3(3x)\sim_0(3x)^3=27x^3$, and ultimately we obtain the equivalence $$\frac{\ln(1+\sin(2x^5))}{\sin(2x^5)}\frac{\sin(2x^5)}{\tan^3(3x)} \cdot \frac{x^2}{(5^{x^{2}} - 1)}\cdot\frac1{x^2}\sim_0\frac{2\not x^5}{27\not x^3\cdot\ln5\cdot \not x^2}=\frac 2{27\ln 5}.$$

Answer 2

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

$$1\cdot1\cdot1^3\cdot1\cdot{2\cdot1^3\over27\ln5}={2\over27\ln5}$$