Limit $\lim_{x\to0^-}{(1+\tan(9x))^{\frac{1}{\arcsin(5x)}}}$

Question

I have a limit: $$\lim_{x\to0^-}{(1+\tan(9x))^{\frac{1}{\arcsin(5x)}}}$$

Are these steps correct?

1. Substitution: $x = n$, $n\to\infty$:

$$\lim_{n\to\infty}=(1+\frac{\sin(9n)}{\cos(9n)})^{\frac{1}{\arcsin(5n)}}$$ $$\lim_{n\to\infty}=e^{\frac{\sin(9n)}{\cos(9n)\arcsin(5n)}}$$

2. Back from substitution: $n = x$, $x\to0^-$:

$$\lim_{x\to0^-}=e^{\frac{\sin(9x)}{\cos(9x)\arcsin(5x)}}$$ $$\lim_{x\to0^-}=e^{\frac{\sin(9x)}{\cos(9n)\arcsin(5x)}\cdot\frac{9x5x}{9x5x}}$$ $$\lim_{x\to0^-}=e^{\frac{9x}{5x\cos(9x)}}$$ $$\lim_{x\to0^-}=e^{\frac{9}{5\cos(9\cdot0^-)}}$$ $$\lim_{x\to0^-}=e^{\frac{9}{5}}$$

Answer 1

First, it is enough to find the limit of the logarithm of the expression, i. e. of $\;\dfrac1{\arcsin5x}\ln\bigl(1+\tan 9x\bigr)$.

Now use equivalents:

• $\ln(1+u)\sim_0u$, hence $\;\ln(1+\tan9x)\sim_0\tan9x\sim_0 9x $;
• $\arcsin 5x\sim_0 5x$.

Thus $$\dfrac1{\arcsin5x}\ln\bigl(1+\tan 9x\bigr)\sim_0 \frac1{5x}\cdot9x=\frac95, $$ whence $$\lim_{x\to 0}(1+\tan9x)^\tfrac1{\arcsin5x}=\mathrm e^{\tfrac95}.$$

Answer 2

Several solid answers have already been posted using a variety of calculus-based techniques. I thought that it would be instructive to supplement these answers with one that only relies on inequalities that can be obtained without calculus. To that end, we proceed.

We begin by writing the limit of interest as

$$\lim_{x\to 0^-}\left(1+\tan(9x)\right)^{1/\arcsin(5x)}=\lim_{x\to 0^-}\exp\left(\frac{\log\left(1+\tan(9x)\right)}{\arcsin(5x)}\right)$$

Now, using standard tools only, I showed in THIS ANSWER that the log function satisfies the inequalities

$$\frac{x}{1+x}\le \log (1+x)\le x \tag 1$$

for $x>-1$.

And using elementary geometry, it can be shown that for $-\pi/2< x<0$

$$x \le \sin x \le x\cos x \tag 2$$

Next, we see from $(2)$ that for $-\pi/2 < x<0$

$$ \frac{x}{\cos x} \le \tan x \le x \tag 3$$

and for $-1<x<0$

$$\frac{x}{\sqrt{1-x^2}} \le \arcsin (x) \le x \tag 4$$

Therefore, we have for $-x_0 < x < 0$, where $x_0\approx 0.0821205659635873 $ is the root of $9x_0+\cos (9x_0)=0$

$$ \frac{9x}{5x} \le \frac{\log\left(1+\tan(9x)\right)}{\arcsin(5x)} \le \frac{\frac{9x}{9x+\cos (9x)}}{\frac{5x}{\sqrt{1-(5x)^2}}}$$

By the squeeze theorem

$$\lim_{x\to 0^-} \frac{\log\left(1+\tan(9x)\right)}{\arcsin(5x)} =\frac95$$

and inasmuch as the exponential function is continuous, we have

$$\lim_{x\to 0^-}\left(1+\tan(9x)\right)^{1/\arcsin(5x)}=e^{9/5}$$

Answer 3

Your first step is wrong in that you cannot put x=n with x going to 0 and also n going to infinity.If you meant x=1/n then your 1st and 2nd equations are wrong as 1/n should appear everywhere you have n. And your 3rd is true but requires proof. Your mathematical grammar needs polishing too. For $x$ close enough to but not equal to $0$ let $$f (x)=(1+\tan 9 x)^{1/\arcsin 5 x}.$$ $$\text {We have } 1+\tan 9 x=(1+ 9 x)(1+x h(x))$$ $$\text {where } \lim_{x\to 0}h(x)=0.$$ $$ \text {We have } 1/\arcsin 5 x= (1/5 x)(1+xk(x)) \text { where } \lim_{x\to 0}k(x)=0.$$Therefore $f(x)=((1+ 9 x)^{1/5 x})^{[1+x k(x)]}p(x) $ where $$p(x)=[((1+x h(x))^{1/5 x}]^{[1+x k(x)]}.$$ Your first equation involving $e$ is valid only because $(1+\tan 9 x) /(1+9 x)$ and $(\arcsin 5 x)/5 x$ converge to $1$ (as $x\to 0^-$) sufficiently fast that $\lim_{x\to 0^-}p(x)=1.$ And since $\lim_{z\to 0, z\ne 0} (1+z)^{1/z}=e$, we obtain $f(x)\to \exp (5/9)$ as $x\to 0^-$.....To show that $p(x)$ does go to $1$ consider that for $x\ne 0$ but close enough to $0$ we have $h(x)\ne 0 $ and $\lim_{x\to 0^-}x h(x)=0$. So $$p(x)= [y(x)]^{[1+x k(x)]} \text { where } y(x)=[(1+ x h(x)]^{1/x h(x)}]^{h(x)/5}, \text { and}$$ $$\lim_{x\to 0}(1+xh(x))^{1/x h(x)}=e.$$ And the exponent $h(x)/5$ goes to $0$, and the exponent $1+xk(x)$ goes to $1$...... As a previous answer points out, taking the logarithm of $f(x)$ at the start can simplify this Q considerably.