Final Answer
Step-by-step Solution
Problem to solve:
Solving method
Rewrite the limit using the identity: $a^x=e^{x\ln\left(a\right)}$
Apply the power rule of limits: $\displaystyle{\lim_{x\to a}f(x)^{g(x)} = \lim_{x\to a}f(x)^{\displaystyle\lim_{x\to a}g(x)}}$
The limit of a constant is just the constant
Rewrite the product inside the limit as a fraction
Plug in the value $\infty $ into the limit
Any expression divided by infinity is equal to zero
Calculating the natural logarithm of $1$
Any expression divided by infinity is equal to zero
If we directly evaluate the limit $\lim_{x\to \infty }\left(\frac{\ln\left(1+\frac{2}{x}\right)}{\frac{1}{x}}\right)$ as $x$ tends to $\infty $, we can see that it gives us an indeterminate form
We can solve this limit by applying L'Hôpital's rule, which consists of calculating the derivative of both the numerator and the denominator separately
Find the derivative of the numerator
The derivative of the natural logarithm of a function is equal to the derivative of the function divided by that function. If $f(x)=ln\:a$ (where $a$ is a function of $x$), then $\displaystyle f'(x)=\frac{a'}{a}$
The derivative of a sum of two functions is the sum of the derivatives of each function
The derivative of the constant function ($1$) is equal to zero
Apply the quotient rule for differentiation, which states that if $f(x)$ and $g(x)$ are functions and $h(x)$ is the function defined by ${\displaystyle h(x) = \frac{f(x)}{g(x)}}$, where ${g(x) \neq 0}$, then ${\displaystyle h'(x) = \frac{f'(x) \cdot g(x) - g'(x) \cdot f(x)}{g(x)^2}}$
The derivative of the constant function ($2$) is equal to zero
The derivative of the linear function is equal to $1$
Combine $1+\frac{2}{x}$ in a single fraction
Multiplying fractions $\frac{x}{2+x} \times \frac{-2}{x^2}$
Simplify the fraction by $x$
Find the derivative of the denominator
Apply the quotient rule for differentiation, which states that if $f(x)$ and $g(x)$ are functions and $h(x)$ is the function defined by ${\displaystyle h(x) = \frac{f(x)}{g(x)}}$, where ${g(x) \neq 0}$, then ${\displaystyle h'(x) = \frac{f'(x) \cdot g(x) - g'(x) \cdot f(x)}{g(x)^2}}$
The derivative of the constant function ($1$) is equal to zero
The derivative of the linear function is equal to $1$
Divide fractions $\frac{\frac{-2}{x\left(2+x\right)}}{\frac{-1}{x^2}}$ with Keep, Change, Flip: $\frac{a}{b}\div c=\frac{a}{b}\div\frac{c}{1}=\frac{a}{b}\times\frac{1}{c}=\frac{a}{b\cdot c}$
Simplify the fraction by $x$
After deriving both the numerator and denominator, the limit results in
Plug in the value $\infty $ into the limit
Any expression multiplied by infinity tends to infinity
Infinity plus any algebraic expression is equal to infinity
If we directly evaluate the limit $\lim_{x\to \infty }\left(\frac{2x}{2+x}\right)$ as $x$ tends to $\infty $, we can see that it gives us an indeterminate form
We can solve this limit by applying L'Hôpital's rule, which consists of calculating the derivative of both the numerator and the denominator separately
Find the derivative of the numerator
The derivative of the linear function times a constant, is equal to the constant
Find the derivative of the denominator
The derivative of a sum of two functions is the sum of the derivatives of each function
The derivative of the constant function ($2$) is equal to zero
The derivative of the linear function is equal to $1$
Divide $2$ by $1$
After deriving both the numerator and denominator, the limit results in
The limit of a constant is just the constant