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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?

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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.

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  • $\begingroup$ 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. $\endgroup$ – cris14 Feb 11 at 21:25
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    $\begingroup$ 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 $\frac{1}{(4^{\tan(x)}+\cos(x))/2} \frac{4^{\tan(x)}ln(4)\sec^2(x) - \sin(x)}{2}$. $\endgroup$ – Bhuvanesh Sundar Feb 11 at 21:39
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    $\begingroup$ 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 $1$. $\endgroup$ – Bhuvanesh Sundar Feb 11 at 21:40
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Call your limit $L$ so, by L'Hôpital's rule, $$\ln L=\lim_{x\to 0}\cos^2 x\frac{\ln 4\cdot 4^{\tan x}\sec^2 x-\sin x}{4^{\tan x}+\cos x}=\frac{\ln 4}{2}=\ln 2\implies L=2.$$

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