Question

Find the limit. Use l’Hospital’s Rule where appropriate. If there is a more elementary method, consider using it. If l’Hospital’s Rule doesn’t apply, explain why.

38. \(\mathop {\lim }\limits_{x \to \infty } {\left( {{e^x} + x} \right)^{1/x}}\)

Short answer

The limit \(\mathop {\lim }\limits_{x \to \infty } {\left( {{e^x} + x} \right)^{1/x}}\) is \(e\).

Step-by-step solution

Given information

The function is \(\mathop {\lim }\limits_{x \to \infty } {\left( {{e^x} + x} \right)^{1/x}}\).

Concept of L Hospital’s Rule

If two functions in the neighborhood of a given point have an infinite limit or zero as a limit and are both differentiable.

If this limit exists, the limit of the quotient of the functions is equal to the limit of the quotient of their derivatives.

Find the limit

Consider, \(y = \mathop {\lim }\limits_{x \to \infty } {\left( {{e^x} + x} \right)^{1/x}}\).

Take natural log of both sides as, \(\ln y = \ln \left( {\mathop {\lim }\limits_{x \to \infty } {{\left( {{e^x} + x} \right)}^{1/x}}} \right)\).

Remember that: \(\mathop {\lim }\limits_{x \to a} f(g(x)) = f\left( {\mathop {\lim }\limits_{x \to a} g(x)} \right)\)

\(\ln y = \mathop {\lim }\limits_{x \to \infty } \ln {\left( {{e^x} + x} \right)^{1/x}}\)

Recall that: \(\ln {x^a} = a\ln x\)

\(\begin{array}{c}\ln y = \mathop {\lim }\limits_{x \to \infty } \frac{1}{x}\ln \left( {{e^x} + x} \right)\\\ln y = \mathop {\lim }\limits_{x \to \infty } \frac{{\ln \left( {{e^x} + x} \right)}}{x}\end{array}\)

Apply L Hopital’s rule

Since the Limit is of the form \(\frac{\infty }{\infty }\), apply L Hospital's Rule.

\(\begin{array}{l}\ln y = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{{{e^x} + 1}}{{{e^x} + x}}}}{1}\\\ln y = \mathop {\lim }\limits_{x \to \infty } \frac{{{e^x} + 1}}{{{e^x} + x}}\end{array}\)

Again, apply L Hospital's Rule as shown below.

\(\begin{array}{c}\ln y = \mathop {\lim }\limits_{x \to \infty } \frac{{{e^x}}}{{{e^x} + 1}}\\Again,{\rm{ }}apply{\rm{ }}L{\rm{ }}Hospital's{\rm{ }}Rule{\rm{ }}as{\rm{ }}shown{\rm{ }}below.\\\ln y = \mathop {\lim }\limits_{x \to \infty } \frac{{{e^x}}}{{{e^x}}}\\ = 1\\{e^{\ln y}} = {e^1}\end{array}\)

Therefore, \(y = e\).