$\lim_{x\to 0}\frac{(e^x-1)( \ln(1+\sin3x)}{\ln(1-\sin x)}$

Question

I need to find the limit of $\dfrac{(e^x-1)(\ln(1+\sin3x))}{\ln(1-\sin x)}$, I know there is equivalences for ($e^x-1) \approx x$ and $\ln(1+x) \approx x$ but I don't know how to change the bottom one. Any hints?

Answer 1

A necessary skill and prerequisite to solve limit is the knowledge of the standard limits; in this case we are referring to (as $x \to 0$)

• $\frac{e^x-1}{x}=1$
• $\frac{\sin x}{x}=1$
• $\frac{\log(1+x)}{x}=1$

using these results we have that

$$\frac{(e^x-1)(\ln(1+\sin(3x)))}{\ln(1-\sin x)}= $$$$=\frac{e^x-1}x\cdot\frac{\ln(1+\sin(3x))}{\sin (3x)} \cdot\frac{-\sin x}{\ln(1-\sin x)}\cdot\frac{\sin(3x)}{3x}\cdot\frac{x}{-\sin x}\cdot 3x \to$$ $$\to1\cdot1\cdot1\cdot1\cdot (-1) \cdot 0=0$$

As an alternative, we can use the first order approximation obtained directly by the standard limits and notably

• $e^x-1=x+o(x)$
• $\sin x= x+o(x)$
• $\log(1+x)=x+o(x)$

to obtain

$$\frac{(e^x-1)(\ln(1+\sin(3x)))}{\ln(1-\sin x)}=\frac{(x+o(x))(\ln(1+3x+o(x))}{\ln(1-x+o(x))}=$$

$$=\frac{(x+o(x))(3x+o(x))}{(-x+o(x))}=\frac{3x^2+o(x^2)}{-x+o(x)}=\frac{3x+o(x)}{-1+o(1)} \to 0$$

As indicated also in other answers, we can also proceed by "asymptotic equivalents" directly obtained by the same satndard limits and notably

• $e^x-1\sim x$
• $\sin x\sim x$
• $\log(1+x)\sim x$

to obtain

$$\frac{(e^x-1)(\ln(1+\sin(3x)))}{\ln(1-\sin x)}\sim \frac{x\cdot \ln(1+3x)}{\ln(1-x)}\sim \frac{x\cdot 3x}{-x}=-3x \to 0$$

which is a faster method but we need to proceed carefully with that since it can lead to great mistakes as for example for $\frac{\sin x-x}{x^3}$ using asymptotic equivalents we obtain

$$\frac{\sin x-x}{x^3}\sim \frac{x-x}{x^3}=0$$

which is wrong.

To see why it is wrong it suffices to use first order approximation to obtain

$$\frac{\sin x-x}{x^3}=\frac{x-x+o(x)}{x^3}=\frac{o(1)}{x^2}$$

which is an indeterminate form and therefore we can't conclude.

To solve it we need others method, notably L'Hospital, Taylor's, etc.

Answer 2

Bear in mind both of your approximations come from the Taylor series expansions of these. For $x$ small, e.g. "near $0$," informally, you can ignore any terms with $x^2$ or higher degrees under the informal rationale that "$x$ is already small and less than $1$, so $x^n$ for $n>1$ is negligibly small. "

In total, you'll want to reference these series:

$$e^x = \sum_{n=0}^\infty \frac{x^n}{n!} \;\;\;\;\; \ln(1-x) = -\sum_{n=1}^\infty \frac{x^n}{n} \;\;\;\;\; \sin(x) = \sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!}$$

Correspondingly, for $x$ small,

$$e^x \approx 1+x \;\;\;\;\; \ln(1-x) \approx -x \;\;\;\;\; \sin(x) \approx x$$

The middle of course gives you $\ln(1+x) \approx x$ equivalently. (You sometimes see slightly different but equivalent series for the natural logarithm, I just used the one that came to mind for me.)

I'll also note that Taylor series aren't the only way to derive these approximations - indeed, it's overkill, since you only need the first derivative, but I feel it's easiest to remember via Taylor series. In any event, to approximate $f$ linearly about the point $x=a$, the general formula is $f(x) \approx f'(a)(x-a) + f(a)$. Take $a=0$ and $f$ to be any of those functions above, and you can derive the approximation that way.

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Anyhow, try those substitutions. In particular, for the denominator, apply the sine one, and then apply the natural logarithm one.

However, note that this isn't the most rigorous way to approach the limits. There's a sort of rigor underlying the use and substitution of these approximations, which I'm skipping over since you seem to be okay with these. To me this suggests that this is okay for your coursework as is.

Answer 3

The fastest is using asymptotic equivalents:

we know that $\;\sin u, \:\ln(1+u)\sim_0 u$ and, as $\mathrm e^x-1=1+x+o(x)-1$ $=x+o(x)$, we have $\:\mathrm e^x-1\sim_0 x$. We obtain $$\frac{(e^x-1)( \ln(1+\sin3x)}{\ln(1-\sin x)}\sim_0\frac{x\cdot 3x}{-x}=-3x\to 0 \quad\text{ as }x\to 0$$