Evaluate the following limits: 1. $\lim_{x \to 0^+} \coth^{-1}(\log_{\frac{7}{9}} x)$ 2. $\lim_{x \to \infty} e^{1-\tanh x}$

Understand the Problem

The question asks to evaluate two limits. The first limit involves the inverse hyperbolic cotangent function and the logarithm function as x approaches 0 from the right. The second limit involves an exponential function with a hyperbolic tangent function in the exponent, as x approaches infinity. We need to apply limit laws and properties of hyperbolic functions to solve these. The user seems to be asking if their solution is correct.

Answer

1. $\lim_{x \to 0^+} \coth^{-1}(\log_{\frac{7}{9}} x) = 0$ 2. $\lim_{x \to \infty} e^{1-\tanh x} = 1$

Steps to Solve

Evaluate the inner limit for the first expression

We need to find the limit of $\log_{\frac{7}{9}} x$ as $x$ approaches $0$ from the right. Since $\frac{7}{9} < 1$, the logarithm function is decreasing. As $x$ approaches $0$ from the right, $\log_{\frac{7}{9}} x$ approaches $+\infty$. $$ \lim_{x \to 0^+} \log_{\frac{7}{9}} x = +\infty $$

Evaluate the outer limit for the first expression

Now we need to find the limit of $\coth^{-1}(u)$ as $u$ approaches $+\infty$. Recall that $\coth(y) = \frac{e^y + e^{-y}}{e^y - e^{-y}}$. Therefore, $\coth^{-1}(x)$ is the value $y$ such that $\coth(y) = x$. As $y \to 0$, $\coth(y)$ gets very large i.e. $\coth(y) \to \infty$.

$$ \lim_{u \to \infty} \coth^{-1}(u) = 0 $$ So, $\lim_{x \to 0^+} \coth^{-1}(\log_{\frac{7}{9}} x) = 0$.

Evaluate the limit of $\tanh x$ for the second expression

We need to find the limit of $\tanh x$ as $x$ approaches $\infty$. Recall that $\tanh x = \frac{e^x - e^{-x}}{e^x + e^{-x}}$. As $x$ approaches infinity, $\tanh x$ approaches 1. $$ \lim_{x \to \infty} \tanh x = 1 $$

Evaluate the limit of the exponent for the second expression

Now we need to find the limit of $1 - \tanh x$ as $x$ approaches infinity. $$ \lim_{x \to \infty} (1 - \tanh x) = 1 - 1 = 0 $$

Evaluate the outer limit for the second expression

Finally, we need to find the limit of $e^{1-\tanh x}$ as $x$ approaches infinity. Since the exponential function is continuous, we have $$ \lim_{x \to \infty} e^{1 - \tanh x} = e^{\lim_{x \to \infty} (1 - \tanh x)} = e^0 = 1 $$

$\lim_{x \to 0^+} \coth^{-1}(\log_{\frac{7}{9}} x) = 0$
$\lim_{x \to \infty} e^{1-\tanh x} = 1$

More Information

The inverse hyperbolic cotangent function approaches 0 as its argument approaches infinity. The hyperbolic tangent function approaches 1 as its argument approaches infinity.

Tips

A common mistake for the second limit is to incorrectly evaluate $\lim_{x \to \infty} \tanh x$ as $-1$ instead of $1$. Also, the first limit combines multiple concepts which could easily lead to errors if you don't understand or know the formulas for inverse hyperbolic cotangent or logarithm functions.

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