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Evaluate $\displaystyle \large \lim_{x \,\to\, \infty}{\normalsize \dfrac{e^x}{\Big(1+\dfrac{1}{x}\Big)^{x^2}}}$

A natural exponential function in $x$ and a special algebraic function in exponential form in $x$ are formed a rational expression. We have to evaluate the limit of the rational function as $x$ approaches infinity in this limit problem.

$\displaystyle \large \lim_{x \,\to\, \infty}{\normalsize \dfrac{e^{\displaystyle x}}{\Bigg(1+\dfrac{1}{x}\Bigg)^{\displaystyle x^2}}}$

Preparing the Rational Expression for Evaluation

The expressions in the numerator and denominator of the rational function are formed in exponential form. So, we have to use the exponential limit rules for evaluating the limit of the rational function. The expression in the denominator is similar to an exponential limit rule. So, we must prepare the given function into our required form.

At this time, no need to disturb the expression in the numerator but focus on the expression in the denominator. Now, express the expression at the exponent position into factor form.

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, \infty}{\normalsize \dfrac{e^{\displaystyle x}}{\Big(1+\dfrac{1}{x}\Big)^{\displaystyle x \times x}}}$

Now, use the power rule of exponents for expressing the expression in the denominator as the power of an exponential function.

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, \infty}{\normalsize \dfrac{e^{\displaystyle x}}{\Bigg(\Big(1+\dfrac{1}{x}\Big)^{\displaystyle x}\Bigg)^{\displaystyle x}}}$

The exponents of the expressions in both numerator and denominator of the rational expression are same. So, it can be simplified by the power of a quotient rule.

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, \infty}{\Large \Bigg(\normalsize \dfrac{e}{\Bigg(1+\dfrac{1}{x}\Bigg)^{\displaystyle x}}\Large \Bigg)^{\displaystyle \normalsize x}}$

Evaluate the Limit of the Rational function

It is time to start the process for evaluating the limit of the function. It can be initiated by the power rule of limits.

$=\,\,\,$ $\Large \Bigg(\displaystyle \large \lim_{x \,\to\, \infty}{\normalsize \dfrac{e}{\Big(1+\dfrac{1}{x}\Big)^{\displaystyle x}}}\Large \Bigg)^{\displaystyle \large \lim_{x \,\to\, \infty}{\normalsize x}}$

The limit of the rational expression in base position can be evaluated by the quotient rule of limits. It helps us to evaluate the limit of the rational expression by the quotient of the limits of the expressions.

$=\,\,\,$ $\Large \Bigg(\normalsize \dfrac{\displaystyle \large \lim_{x \,\to\, \infty}{\normalsize e}}{\displaystyle \large \lim_{x \,\to\, \infty}{\normalsize \Big(1+\dfrac{1}{x}\Big)^{\displaystyle x}}}\Large \Bigg)^{\displaystyle \large \lim_{x \,\to\, \infty}{\normalsize x}}$

$=\,\,\,$ $\Big(\dfrac{e}{e}\Big)^{\infty}$

$\require{cancel} =\,\,\,$ $\Big(\dfrac{\cancel{e}}{\cancel{e}}\Big)^{\infty}$

$=\,\,\,$ $(1)^{\infty}$

$=\,\,\,$ $1^{\infty}$

It is evaluated that the limit of the given rational function is $1$ raised to the power of infinity. It means that the limit of the rational expression is indeterminate as $x$ approaches infinity. So, we must think for an alternative method for evaluating the limit of the rational function in algebraic form.

Technique to Remove the Indeterminate form

It is fact that the indeterminate form is come due to the formation of the function by the exponential form functions. We can stop the rational function to give the indeterminate form when the function is released from the exponential form expressions. So, it can be done by the logarithmic system.

If $y \,=\, \dfrac{e^x}{\Big(1+\dfrac{1}{x}\Big)^{x^2}}$ then take logarithm both sides of the equation.

$\implies$ $\log_e{(y)}$ $\,=\,$ $\log_e{\Large \Bigg(\normalsize \dfrac{e^{\displaystyle x}}{\Bigg(1+\dfrac{1}{x}\Bigg)^{\displaystyle x^2}}\Large \Bigg)}$

The expression in the right side of the equation can be expanded by the quotient rule of logarithms.

$\implies$ $\log_e{(y)}$ $\,=\,$ $\log_e{\Big(\normalsize e^{\displaystyle x}\Big)}$ $-$ $\log_e{\Bigg(1+\dfrac{1}{x}\Bigg)^{\displaystyle x^2}}$

The exponential form of each term can be removed by the power rule of logarithms.

$\implies$ $\log_e{(y)}$ $\,=\,$ $x \times \log_e{(e)}$ $-$ $x^2 \times \log_e{\Bigg(1+\dfrac{1}{x}\Bigg)}$

The natural logarithm of $e$ is one as per the logarithm rule of base.

$\implies$ $\log_e{(y)}$ $\,=\,$ $x \times 1$ $-$ $x^2 \times \log_e{\Bigg(1+\dfrac{1}{x}\Bigg)}$

$\implies$ $\log_e{(y)}$ $\,=\,$ $x$ $-$ $x^2 \times \log_e{\Bigg(1+\dfrac{1}{x}\Bigg)}$

The logarithm of sum of the terms can be expanded by the logarithmic expansion rule. Now, simplify the expression at the right hand side of the equation.

$\implies$ $\log_e{(y)}$ $\,=\,$ $x$ $-$ $x^2 \times \Bigg(\dfrac{1}{x}$ $-$ $\dfrac{\Big(\dfrac{1}{x}\Big)^2}{2}$ $+$ $\dfrac{\Big(\dfrac{1}{x}\Big)^3}{3}$ $-$ $\dfrac{\Big(\dfrac{1}{x}\Big)^4}{4}$ $+$ $\cdots\Bigg)$

$\implies$ $\log_e{(y)}$ $\,=\,$ $x$ $-$ $x^2 \times \Bigg(\dfrac{1}{x}$ $-$ $\dfrac{\Big(\dfrac{1^2}{x^2}\Big)}{2}$ $+$ $\dfrac{\Big(\dfrac{1^3}{x^3}\Big)}{3}$ $-$ $\dfrac{\Big(\dfrac{1^4}{x^4}\Big)}{4}$ $+$ $\cdots\Bigg)$

$\implies$ $\log_e{(y)}$ $\,=\,$ $x$ $-$ $x^2 \times \Bigg(\dfrac{1}{x}$ $-$ $\dfrac{\Big(\dfrac{1}{x^2}\Big)}{2}$ $+$ $\dfrac{\Big(\dfrac{1}{x^3}\Big)}{3}$ $-$ $\dfrac{\Big(\dfrac{1}{x^4}\Big)}{4}$ $+$ $\cdots\Bigg)$

$\implies$ $\log_e{(y)}$ $\,=\,$ $x$ $-$ $x^2 \times \Bigg(\dfrac{1}{x}$ $-$ $\dfrac{1}{2 \times x^2}$ $+$ $\dfrac{1}{3 \times x^3}$ $-$ $\dfrac{1}{4 \times x^4}$ $+$ $\cdots\Bigg)$

$\implies$ $\log_e{(y)}$ $\,=\,$ $x$ $-$ $\Bigg(\dfrac{x^2 \times 1}{x}$ $-$ $\dfrac{x^2 \times 1}{2 \times x^2}$ $+$ $\dfrac{x^2 \times 1}{3 \times x^3}$ $-$ $\dfrac{x^2 \times 1}{4 \times x^4}$ $+$ $\cdots\Bigg)$

$\implies$ $\log_e{(y)}$ $\,=\,$ $x$ $-$ $\Bigg(\dfrac{\cancel{x^2} \times 1}{\cancel{x}}$ $-$ $\dfrac{\cancel{x^2} \times 1}{2 \times \cancel{x^2}}$ $+$ $\dfrac{\cancel{x^2} \times 1}{3 \times \cancel{x^3}}$ $-$ $\dfrac{\cancel{x^2} \times 1}{4 \times \cancel{x^4}}$ $+$ $\cdots\Bigg)$

$\implies$ $\log_e{(y)}$ $\,=\,$ $x$ $-$ $\Bigg(\dfrac{x \times 1}{1}$ $-$ $\dfrac{1 \times 1}{2 \times 1}$ $+$ $\dfrac{1 \times 1}{3 \times x}$ $-$ $\dfrac{1 \times 1}{4 \times x^2}$ $+$ $\cdots\Bigg)$

$\implies$ $\log_e{(y)}$ $\,=\,$ $x$ $-$ $\Bigg(\dfrac{x}{1}$ $-$ $\dfrac{1}{2}$ $+$ $\dfrac{1}{3x}$ $-$ $\dfrac{1}{4x^2}$ $+$ $\cdots\Bigg)$

$\implies$ $\log_e{(y)}$ $\,=\,$ $x$ $-$ $\Bigg(x$ $-$ $\dfrac{1}{2}$ $+$ $\dfrac{1}{3x}$ $-$ $\dfrac{1}{4x^2}$ $+$ $\cdots\Bigg)$

$\implies$ $\log_e{(y)}$ $\,=\,$ $x$ $-$ $x$ $+$ $\dfrac{1}{2}$ $-$ $\dfrac{1}{3x}$ $+$ $\dfrac{1}{4x^2}$ $-$ $\cdots$

$\implies$ $\log_e{(y)}$ $\,=\,$ $\cancel{x}$ $-$ $\cancel{x}$ $+$ $\dfrac{1}{2}$ $-$ $\dfrac{1}{3x}$ $+$ $\dfrac{1}{4x^2}$ $-$ $\cdots$

$\implies$ $\log_e{(y)}$ $\,=\,$ $\dfrac{1}{2}$ $-$ $\dfrac{1}{3x}$ $+$ $\dfrac{1}{4x^2}$ $-$ $\cdots$

Evaluate the Limit of the Algebraic equation

The simplification of the logarithmic equation is completed. Now, let’s us evaluate the limit of the log equation as $x$ close to infinity.

$\implies$ $\displaystyle \large \lim_{x \,\to\, \infty}{\normalsize \log_e{(y)}}$ $\,=\,$ $\displaystyle \large \lim_{x \,\to\, \infty}{\normalsize \Bigg(\dfrac{1}{2}}$ $-$ $\dfrac{1}{3x}$ $+$ $\dfrac{1}{4x^2}$ $-$ $\cdots\Bigg)$

Now, evaluate the limit of the infinite series as $x$ approaches infinity by the direct substitution method.

$\implies$ $\displaystyle \large \lim_{x \,\to\, \infty}{\normalsize \log_e{(y)}}$ $\,=\,$ $\dfrac{1}{2}$ $-$ $\dfrac{1}{3(\infty)}$ $+$ $\dfrac{1}{4(\infty)^2}$ $-$ $\cdots$

$\implies$ $\displaystyle \large \lim_{x \,\to\, \infty}{\normalsize \log_e{(y)}}$ $\,=\,$ $\dfrac{1}{2}$ $-$ $\dfrac{1}{3 \times \infty}$ $+$ $\dfrac{1}{4 \times \infty^2}$ $-$ $\cdots$

$\implies$ $\displaystyle \large \lim_{x \,\to\, \infty}{\normalsize \log_e{(y)}}$ $\,=\,$ $\dfrac{1}{2}$ $-$ $\dfrac{1}{\infty}$ $+$ $\dfrac{1}{4 \times \infty}$ $-$ $\cdots$

$\implies$ $\displaystyle \large \lim_{x \,\to\, \infty}{\normalsize \log_e{(y)}}$ $\,=\,$ $\dfrac{1}{2}$ $-$ $0$ $+$ $\dfrac{1}{\infty}$ $-$ $\cdots$

$\implies$ $\displaystyle \large \lim_{x \,\to\, \infty}{\normalsize \log_e{(y)}}$ $\,=\,$ $\dfrac{1}{2}$ $+$ $0$ $-$ $\cdots$

The limit of the composition of the functions can be evaluated by the limit rule of composition function.

$\implies$ $\displaystyle \large \lim_{x \,\to\, \infty}{\normalsize \log_e{(y)}}$ $\,=\,$ $\dfrac{1}{2}$

Now, eliminate the logarithm from the function by the relationship between exponents and logarithms.

$\implies$ $\log_e{\Bigg(\displaystyle \large \lim_{x \,\to\, \infty}{\normalsize y\Bigg)}}$ $\,=\,$ $\dfrac{1}{2}$

$\implies$ $\displaystyle \large \lim_{x \,\to\, \infty}{\normalsize y}$ $\,=\,$ $e^\frac{1}{2}$

$\implies$ $\displaystyle \large \lim_{x \,\to\, \infty}{\normalsize y}$ $\,=\,$ $\sqrt{e}$

Actually, we have assumed that $y \,=\, \dfrac{e^x}{\Big(1+\dfrac{1}{x}\Big)^{x^2}}$.

$\,\,\,\therefore\,\,\,\,\,\,$ $\displaystyle \large \lim_{x \,\to\, \infty}{\normalsize \dfrac{e^x}{\Big(1+\dfrac{1}{x}\Big)^{x^2}}}$ $\,=\,$ $\sqrt{e}$

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