Comprehensive Exercises

Idea: This is a $\frac{0}{0}$ indeterminate form, so we can use the factorization method.

Detailed steps:

1. Check the type of indeterminate form:

   • When $x = 1$, the numerator $1 - 3 + 3 - 1 = 0$
   • When $x = 1$, the denominator $1 - 2 + 1 = 0$
   • Hence it is a $\frac{0}{0}$ indeterminate form

2. Factor the numerator: $x^3 - 3x^2 + 3x - 1 = (x - 1)^3$

3. Factor the denominator: $x^2 - 2x + 1 = (x - 1)^2$

4. Cancel the zero factor: $\lim_{x \to 1} \frac{(x - 1)^3}{(x - 1)^2} = \lim_{x \to 1} (x - 1) = 0$

Answer: $\lim_{x \to 1} \frac{x^3 - 3x^2 + 3x - 1}{x^2 - 2x + 1} = 0$

Idea: This is a $\frac{0}{0}$ indeterminate form containing radicals, so we can use the rationalization method.

Detailed steps:

1. Check the type of indeterminate form:

   • As $x \to 0$, the numerator $\sqrt{1 + x} - \sqrt{1 - x} \to 0$
   • As $x \to 0$, the denominator $x \to 0$
   • Hence it is a $\frac{0}{0}$ indeterminate form

2. Rationalize the numerator: $\frac{\sqrt{1 + x} - \sqrt{1 - x}}{x} = \frac{(\sqrt{1 + x} - \sqrt{1 - x})(\sqrt{1 + x} + \sqrt{1 - x})}{x(\sqrt{1 + x} + \sqrt{1 - x})}$

3. Simplify: $\frac{(\sqrt{1 + x} - \sqrt{1 - x})(\sqrt{1 + x} + \sqrt{1 - x})}{x(\sqrt{1 + x} + \sqrt{1 - x})} = \frac{(1 + x) - (1 - x)}{x(\sqrt{1 + x} + \sqrt{1 - x})} = \frac{2x}{x(\sqrt{1 + x} + \sqrt{1 - x})}$

4. Cancel $x$: $\frac{2x}{x(\sqrt{1 + x} + \sqrt{1 - x})} = \frac{2}{\sqrt{1 + x} + \sqrt{1 - x}}$

5. Take the limit: $\lim_{x \to 0} \frac{2}{\sqrt{1 + x} + \sqrt{1 - x}} = \frac{2}{\sqrt{1 + 0} + \sqrt{1 - 0}} = \frac{2}{1 + 1} = 1$

Answer: $\lim_{x \to 0} \frac{\sqrt{1 + x} - \sqrt{1 - x}}{x} = 1$

Idea: This is a $\frac{0}{0}$ indeterminate form, so we can use L’Hôpital’s rule or equivalent infinitesimal substitution.

Detailed steps:

1. Check the type of indeterminate form:

   • As $x \to 0$, the numerator $\sin x - x \to 0$
   • As $x \to 0$, the denominator $x^3 \to 0$
   • Hence it is a $\frac{0}{0}$ indeterminate form

2. Use L’Hôpital’s rule:

   • $(\sin x - x)' = \cos x - 1$
   • $(x^3)' = 3x^2$

3. Take the limit of the derivatives: $\lim_{x \to 0} \frac{\cos x - 1}{3x^2}$ This is still a $\frac{0}{0}$ indeterminate form

4. Use L’Hôpital’s rule again:

   • $(\cos x - 1)' = -\sin x$
   • $(3x^2)' = 6x$

5. Take the limit of the second derivatives: $\lim_{x \to 0} \frac{-\sin x}{6x} = -\frac{1}{6} \lim_{x \to 0} \frac{\sin x}{x} = -\frac{1}{6}$

Answer: $\lim_{x \to 0} \frac{\sin x - x}{x^3} = -\frac{1}{6}$

Idea: This is a $\frac{\infty}{\infty}$ indeterminate form, so we can use L’Hôpital’s rule or directly compare the highest-degree terms.

Detailed steps:

1. Check the type of indeterminate form:

   • As $x \to \infty$, both the numerator and the denominator tend to infinity
   • Hence it is a $\frac{\infty}{\infty}$ indeterminate form

2. Method 1: directly compare the highest-degree terms $\lim_{x \to \infty} \frac{x^3 + 2x^2 + 3x + 4}{x^3 + x^2 + x + 1} = \lim_{x \to \infty} \frac{x^3(1 + \frac{2}{x} + \frac{3}{x^2} + \frac{4}{x^3})}{x^3(1 + \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3})} = \lim_{x \to \infty} \frac{1 + \frac{2}{x} + \frac{3}{x^2} + \frac{4}{x^3}}{1 + \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3}} = 1$

3. Method 2: use L’Hôpital’s rule

   • Derivative of the numerator: $3x^2 + 4x + 3$
   • Derivative of the denominator: $3x^2 + 2x + 1$
   • Still a $\frac{\infty}{\infty}$ indeterminate form
   • Differentiate again: numerator $6x + 4$, denominator $6x + 2$
   • Still a $\frac{\infty}{\infty}$ indeterminate form
   • Differentiate again: numerator $6$, denominator $6$
   • The final limit is $1$

Answer: $\lim_{x \to \infty} \frac{x^3 + 2x^2 + 3x + 4}{x^3 + x^2 + x + 1} = 1$

Idea: This limit is hard to compute directly, so we can use the squeeze theorem.

Detailed steps:

1. Analyze the features of the function:

   • As $x \to 0$, we have $x^2 \to 0$
   • $\sin \frac{1}{x}$ oscillates between $[-1, 1]$

2. Construct the inequality:

   • Since $-1 \leq \sin \frac{1}{x} \leq 1$
   • We have $-x^2 \leq x^2 \sin \frac{1}{x} \leq x^2$

3. Find the limits of the boundary functions:

   • $\lim_{x \to 0} (-x^2) = 0$
   • $\lim_{x \to 0} x^2 = 0$

4. Apply the squeeze theorem:

   • Since $-x^2 \leq x^2 \sin \frac{1}{x} \leq x^2$
   • And $\lim_{x \to 0} (-x^2) = \lim_{x \to 0} x^2 = 0$
   • Therefore $\lim_{x \to 0} x^2 \sin \frac{1}{x} = 0$

Answer: $\lim_{x \to 0} x^2 \sin \frac{1}{x} = 0$

Idea: This is a $\frac{0}{0}$ indeterminate form, so we can use L’Hôpital’s rule.

Detailed steps:

1. Check the type of indeterminate form:

   • As $x \to 0$, the numerator $e^x - 1 - x \to 0$
   • As $x \to 0$, the denominator $x^2 \to 0$
   • Hence it is a $\frac{0}{0}$ indeterminate form

2. Use L’Hôpital’s rule:

   • $(e^x - 1 - x)' = e^x - 1$
   • $(x^2)' = 2x$

3. Take the limit of the derivatives: $\lim_{x \to 0} \frac{e^x - 1}{2x}$ This is still a $\frac{0}{0}$ indeterminate form

4. Use L’Hôpital’s rule again:

   • $(e^x - 1)' = e^x$
   • $(2x)' = 2$

5. Take the limit of the second derivatives: $\lim_{x \to 0} \frac{e^x}{2} = \frac{e^0}{2} = \frac{1}{2}$

Answer: $\lim_{x \to 0} \frac{e^x - 1 - x}{x^2} = \frac{1}{2}$