L’Hospital’s Rule and Indeterminate Forms — Question 6

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Question 6

Evaluate the limit: limx→∞(x−lnx)\lim_{x \to \infty} \left( x - \ln x \right)

Original worksheet page 1: question and worked solution for 4-10-006
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Question 6 - Solution

As x→∞x \to \infty, we observe: x→∞,ln⁡x→∞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−ln⁡x=x(1−ln⁡xx)x - \ln x = x \left( 1 - \frac{\ln x}{x} \right)

Now evaluate: limx→∞ln⁡xx=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}

Original worksheet page 2: question and worked solution for 4-10-006

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