Show solutionHide solution+ Question 6 Evaluate the limit: limx→∞(x−lnx)\lim_{x \to \infty} \left( x - \ln x \right) Question 6 - Solution As x→∞x \to \infty, we observe: x→∞,lnx→∞x \to \infty, \quad \ln x \to \infty So the limit is of the form: ∞−∞(indeterminate form)\infty - \infty \quad \text{(indeterminate form)} To apply L’Hospital’s Rule, we rewrite the expression to form a quotient. Factor: x−lnx=x(1−lnxx)x - \ln x = x \left( 1 - \frac{\ln x}{x} \right) Now evaluate: limx→∞lnxx=0(since exponential grows faster)\lim_{x \to \infty} \frac{\ln x}{x} = 0 \quad \text{(since exponential grows faster)} So: limx→∞(x−lnx)=∞⋅(1−0)=∞\lim_{x \to \infty} \left( x - \ln x \right) = \infty \cdot (1 - 0) = \infty Final Answer: ∞\boxed{\infty}