We are tasked with evaluating the following limit:

$\lim_{x \to \infty} \left( 14x e^{\frac{1}{x}} - 14x \right).$

Step-by-step solution:

1. Factor out $14x$: $\lim_{x \to \infty} \left( 14x e^{\frac{1}{x}} - 14x \right) = \lim_{x \to \infty} 14x \left( e^{\frac{1}{x}} - 1 \right).$

2. Expand $e^{\frac{1}{x}}$ for large $x$: Using the Taylor series expansion for $e^u$ around $u = 0$: $e^{\frac{1}{x}} \approx 1 + \frac{1}{x} + \frac{1}{2x^2} + \dots,$ for large $x$, higher-order terms become negligible.

   Substituting this into $e^{\frac{1}{x}} - 1$: $e^{\frac{1}{x}} - 1 \approx \frac{1}{x} + \frac{1}{2x^2}.$

3. Substitute into the limit: $\lim_{x \to \infty} 14x \left( e^{\frac{1}{x}} - 1 \right) = \lim_{x \to \infty} 14x \left( \frac{1}{x} + \frac{1}{2x^2} \right).$

4. Simplify: $\lim_{x \to \infty} 14x \left( \frac{1}{x} + \frac{1}{2x^2} \right) = \lim_{x \to \infty} \left( 14 + \frac{14}{2x} \right).$

5. Take the limit: As $x \to \infty$, the term $\frac{14}{2x}$ goes to 0: $\lim_{x \to \infty} \left( 14 + \frac{14}{2x} \right) = 14.$

Final Answer:

$\lim_{x \to \infty} \left( 14x e^{\frac{1}{x}} - 14x \right) = 14.$

Would you like further details or clarifications?

Related Questions:

1. How does the Taylor expansion help approximate functions near certain values?
2. What happens to the higher-order terms of $e^u$ as $x \to \infty$?
3. How do limits behave with exponential growth rates compared to polynomial growth?
4. Can you compute similar limits with logarithmic functions involved?
5. How would this calculation differ if the term were $e^{-1/x}$?

Tip:

When evaluating limits with exponential or logarithmic functions, approximations via Taylor expansions or series expansions are incredibly powerful tools.