Question

Let $\alpha = \lim_{x \to 0} (\frac{e^x-1}{x})^{\frac{1}{x}}$, $\beta = \lim_{x \to 1} (\ln(ex))^{\log_x e}$ and $\gamma = \lim_{n \to \infty} \frac{\sqrt{n} \sin(5^n) + \cos^2(2^n)}{n+5}$, then $\frac{\alpha^2 + \beta + \gamma}{e}$ is _______.

Answer 1

2

Answer 2

To evaluate the expression $\frac{\alpha^2 + \beta + \gamma}{e}$, we need to calculate the values of $\alpha$, $\beta$, and $\gamma$ separately.

1. Calculate $\alpha = \lim_{x \to 0} \left(\frac{e^x-1}{x}\right)^{\frac{1}{x}}$

This limit is of the indeterminate form $1^\infty$. We can evaluate it using the formula $\lim_{x \to a} [f(x)]^{g(x)} = e^{\lim_{x \to a} g(x)[f(x)-1]}$ if $\lim_{x \to a} f(x) = 1$ and $\lim_{x \to a} g(x) = \infty$. Here, $f(x) = \frac{e^x-1}{x}$ and $g(x) = \frac{1}{x}$. As $x \to 0$, $\lim_{x \to 0} \frac{e^x-1}{x} = 1$ (standard limit) and $\lim_{x \to 0} \frac{1}{x} = \infty$. So, $\alpha = e^{\lim_{x \to 0} \frac{1}{x} \left(\frac{e^x-1}{x} - 1\right)}$. Let's evaluate the exponent $L_1 = \lim_{x \to 0} \frac{1}{x} \left(\frac{e^x-1-x}{x}\right) = \lim_{x \to 0} \frac{e^x-1-x}{x^2}$. This is of the form $\frac{0}{0}$. We can use L'Hopital's Rule twice or series expansion. Using series expansion: $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + O(x^4)$. So, $e^x-1-x = (1 + x + \frac{x^2}{2} + \frac{x^3}{6} + O(x^4)) - 1 - x = \frac{x^2}{2} + \frac{x^3}{6} + O(x^4)$. $L_1 = \lim_{x \to 0} \frac{\frac{x^2}{2} + \frac{x^3}{6} + O(x^4)}{x^2} = \lim_{x \to 0} \left(\frac{1}{2} + \frac{x}{6} + O(x^2)\right) = \frac{1}{2}$. Therefore, $\alpha = e^{1/2} = \sqrt{e}$.

2. Calculate $\beta = \lim_{x \to 1} (\ln(ex))^{\log_x e}$

This limit is also of the indeterminate form $1^\infty$. As $x \to 1$, $\ln(ex) \to \ln(e \cdot 1) = \ln e = 1$. And $\log_x e = \frac{\ln e}{\ln x} = \frac{1}{\ln x}$. As $x \to 1$, $\ln x \to 0$, so $\frac{1}{\ln x} \to \infty$. Let $f(x) = \ln(ex)$ and $g(x) = \log_x e$. $\beta = e^{\lim_{x \to 1} \log_x e (\ln(\ln(ex)))}$. Let's evaluate the exponent $L_2 = \lim_{x \to 1} \frac{1}{\ln x} \ln(\ln(ex))$. We know $\ln(ex) = \ln e + \ln x = 1 + \ln x$. So, $L_2 = \lim_{x \to 1} \frac{\ln(1+\ln x)}{\ln x}$. Let $y = \ln x$. As $x \to 1$, $y \to 0$. $L_2 = \lim_{y \to 0} \frac{\ln(1+y)}{y}$. This is a standard limit, which equals 1. Therefore, $\beta = e^1 = e$.

3. Calculate $\gamma = \lim_{n \to \infty} \frac{\sqrt{n} \sin(5^n) + \cos^2(2^n)}{n+5}$

We use the Squeeze Theorem for this limit. We know that $-1 \le \sin(5^n) \le 1$ and $0 \le \cos^2(2^n) \le 1$. So, for the numerator: $-\sqrt{n} \cdot 1 + 0 \le \sqrt{n} \sin(5^n) + \cos^2(2^n) \le \sqrt{n} \cdot 1 + 1$. $-\sqrt{n} \le \sqrt{n} \sin(5^n) + \cos^2(2^n) \le \sqrt{n}+1$. Now, divide by $n+5$ (which is positive for large $n$): $\frac{-\sqrt{n}}{n+5} \le \frac{\sqrt{n} \sin(5^n) + \cos^2(2^n)}{n+5} \le \frac{\sqrt{n}+1}{n+5}$. Let's find the limits of the lower and upper bounds as $n \to \infty$: $\lim_{n \to \infty} \frac{-\sqrt{n}}{n+5} = \lim_{n \to \infty} \frac{-1/\sqrt{n}}{1+5/n} = \frac{0}{1+0} = 0$. $\lim_{n \to \infty} \frac{\sqrt{n}+1}{n+5} = \lim_{n \to \infty} \frac{1/\sqrt{n}+1/n}{1+5/n} = \frac{0+0}{1+0} = 0$. By the Squeeze Theorem, $\gamma = 0$.

4. Calculate $\frac{\alpha^2 + \beta + \gamma}{e}$

Substitute the values of $\alpha$, $\beta$, and $\gamma$: $\alpha^2 = (\sqrt{e})^2 = e$. $\frac{\alpha^2 + \beta + \gamma}{e} = \frac{e + e + 0}{e} = \frac{2e}{e} = 2$.

The final answer is $\boxed{2}$.

Explanation of the solution:

1. Calculate $\alpha$: The limit is of $1^\infty$ form. Use the property $\lim [f(x)]^{g(x)} = e^{\lim g(x)(f(x)-1)}$. Evaluate the exponent $\lim_{x \to 0} \frac{e^x-1-x}{x^2}$ using Taylor series expansion of $e^x$ around $x=0$ up to $x^2$ term. This yields $\frac{1}{2}$. So $\alpha = e^{1/2} = \sqrt{e}$.
2. Calculate $\beta$: The limit is of $1^\infty$ form. Use the same property as for $\alpha$. The exponent becomes $\lim_{x \to 1} \frac{\ln(\ln(ex))}{\ln x}$. Substitute $\ln(ex) = 1 + \ln x$. Let $y = \ln x$. The exponent simplifies to $\lim_{y \to 0} \frac{\ln(1+y)}{y}$, which is a standard limit equal to 1. So $\beta = e^1 = e$.
3. Calculate $\gamma$: Use the Squeeze Theorem. The terms $\sin(5^n)$ and $\cos^2(2^n)$ are bounded between $-1$ and $1$, and $0$ and $1$ respectively. This gives bounds for the numerator: $-\sqrt{n} \le \text{Numerator} \le \sqrt{n}+1$. Divide by $n+5$ and take limits. Both lower and upper bounds approach 0 as $n \to \infty$. Thus, $\gamma = 0$.
4. Final Calculation: Substitute $\alpha^2 = e$, $\beta = e$, and $\gamma = 0$ into the expression $\frac{\alpha^2 + \beta + \gamma}{e}$. This gives $\frac{e+e+0}{e} = \frac{2e}{e} = 2.