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