# Evaluate: $\lim_{x\to0}\left (\frac{4^{\tan(x)}+ \cos(x)}{2}\right)^{\cot(x)}$

I have limit:

$$\lim_{x\to0}\left (\frac{4^{\tan(x)}+ \cos(x)}{2}\right)^{\cot(x)}$$

I tried to use the natural log:

$$\lim_{x\to 0} e^{\dfrac{\ln\left(\dfrac{4^{\tan(x)}+ \cos(x)}{2}\right)}{{\tan(x)}}}$$ But I am stuck from here, I tried multiple approaches but could not find the right result which should be $$2$$

How should I approach this limit?

Use L'Hospital's rule for the exponent ln(...) only. According to this rule, you differentiate the numerator and denominator. Doing so, you get (for the exponent): $$\lim_{x\rightarrow0} \frac{\frac{ (4^{\tan(x)}ln(4)-\sin(x))/2 }{(4^{\tan(x)}+\cos(x))/2}}{\sec^2x} = \frac{\frac{ (4^0ln(4)-0)/2 }{(4^0+1)/2}}{1} = ln(2).$$ Since the limiting value of the exponent is ln(2), the limiting value of your expression is 2.

• would you be so kind and explained me how you find the derivatives, I am trying to reproduce your approach but will still get something different. Commented Feb 11, 2019 at 21:25
• I want to calculate $\frac{ d(numerator)/dx}{d(denominator)/dx}$. Step1: $d\tan(x)/dx = \sec^2(x)$. Step 2: The derivative of $4^{\tan(x)}$ is, by chain rule, $4^{\tan(x)}ln(4)\frac{d(\tan(x))}{dx} = 4^{\tan(x)}ln(4)\sec^2(x)$. The derivative of $\cos(x)$ is $-\sin(x)$. The derivative of $ln(...)$ is $\frac{1}{...}\frac{d(...)}{dx}$. Therefore, by chain rule, the derivative of $ln\left(\frac{4^{\tan(x)}+\cos(x))}{2}\right)$ is $\frac{1}{(4^{\tan(x)}+\cos(x))/2} \frac{4^{\tan(x)}ln(4)\sec^2(x) - \sin(x)}{2}$. Commented Feb 11, 2019 at 21:39
• Step 3: At $x=0$, the derivative of the numerator evaluates to $\frac{1}{(4^0+1)/2} \frac{4^0ln(4) - 0}{2} = ln(4)/2 = ln(2)$. The derivative of the denominator evaluates to $1$. Commented Feb 11, 2019 at 21:40

Call your limit $$L$$ so, by L'Hôpital's rule, $$\ln L=\lim_{x\to 0}\cos^2 x\frac{\ln 4\cdot 4^{\tan x}\sec^2 x-\sin x}{4^{\tan x}+\cos x}=\frac{\ln 4}{2}=\ln 2\implies L=2.$$

L'Hospital isn't necessary at all.

Revision: (source Continuity & Limits/Neprekidnost i limes)

Let: $$\lim_{x\to c}f(x)=1\;\&\;\lim_{x\to c} g(x)=+\infty$$ Then:\begin{aligned}\lim_{x\to c}f(x)^{g(x)}&=\lim_{x\to c}\left(1+(f(x)-1)\right)^{g(x)}\\&=\lim_{x\to c}\left[\left(1+(f(x)-1)\right)^{\frac1{f(x)-1}}\right]^{(f(x)-1)g(x)}\\&=e^{\space\lim\limits_{x\to c}\;(f(x)-1)g(x)}\end{aligned}

Using the standard limits from the table:

$$\lim_{x\to 0}\frac{a^x -1}{x}=\ln a, \;\mathbb R\ni a>0\;\;\&\;\lim_{x\to 0}\frac{\sin x}{x}=1$$

Let's do the following: $$f(x):=\frac{4^{\tan x}+\cos x}2\quad\&\quad g(x)=\cot x$$ \begin{aligned}\implies f(x)-1&=\frac{4^{\tan x}+\cos x}2-1\\&=\frac{4^{\tan x}-1-(1-\cos x)}2.\end{aligned} $$f(x)-1=\frac12\left(\frac{4^{\tan x}-1}{\tan x}\cdot\tan x-\frac{1-\cos x}{x^2}\cdot x^2\right).$$ Hence, \begin{aligned}(f(x)-1)g(x)&=\frac12\left(\frac{4^{\tan x}-1}{\tan x}\cdot 1-\frac{1-\cos x}{x^2}\cdot\frac1{\frac{\sin x}{x}}\cdot x\cos x\right)\\\implies\lim_{x\to 0}(f(x)-1)g(x)&=\frac12\left(\ln 4\cdot1-\frac12\cdot 1\cdot 0\cdot 1\right)\\&=\frac12\cdot\ln 4=\ln2.\\\\\implies L=&\lim_{x\to 0}\left(\frac{4^{\tan x}+\cos x}2\right)^{\cot x}\\&=e^{\ln 2}=2.\end{aligned}

• Nice proof, I like it (+1). Commented Nov 20, 2020 at 12:06
• Thank you, @Angelo ! Commented Nov 21, 2020 at 0:29