evaluate the limit using lhospitals rule lim x→∞ 9xe^(1/x) - 9x

Explanation:

Step1: Rewrite the expression

First, factor out (9x): (9x(e^{\frac{1}{x}} - 1)). Let (t=\frac{1}{x}), as (x\rightarrow\infty), (t\rightarrow0). The expression becomes (\frac{9(e^{t}-1)}{t}).

Step2: Apply L'Hopital's rule

The limit (\lim_{t\rightarrow0}\frac{9(e^{t}-1)}{t}) is in the (\frac{0}{0}) - form. Differentiate the numerator and denominator. The derivative of (9(e^{t}-1)) with respect to (t) is (9e^{t}), and the derivative of (t) with respect to (t) is (1).

Step3: Evaluate the new - limit

(\lim_{t\rightarrow0}\frac{9e^{t}}{1}=9e^{0}).

Answer:

9