L2 - Limits (Limit at infinity and Trig Limit) Review 4

Limit of a Radical Expression

• Expression:[ s = \sqrt{x^2 + 2x - 1} - \sqrt{x^2 - 1} ]

• Objective: Find the limit as ( x ) approaches negative infinity.

  • The function involves radical expressions.

Conceptual Approach

• As ( x \to -\infty ), intuitively, the two square roots might appear to cancel each other due to similar leading terms (both of degree one).

• Direct substitution is not possible due to infinity.

Common Factoring Technique

• Factor out ( \sqrt{x^2} ) from each square root term: [ \sqrt{x^2(1 + \frac{2x}{x^2} - \frac{1}{x^2})} - \sqrt{x^2(1 - \frac{1}{x^2})} ]

• This leads to: [ \lim_{x \to -\infty} \left( \sqrt{x^2} \left( \sqrt{1 + \frac{2}{x} - \frac{1}{x^2}} - \sqrt{1 - \frac{1}{x^2}} \right) \right) ]

Absolute Value Assessment

• Recognizing that ( \sqrt{x^2} = |x| ) leads to:

  • Since ( x ) is negative, ( |x| = -x ):[ \lim_{x \to -\infty} -x \left( \sqrt{1 + 0} - \sqrt{1 - 0} \right) ]

Indeterminate Form

• This results in ( 0 - 0 ), which is an indeterminate form of ( \infty \cdot 0 ).

• Determining which part dominates is unclear, leading to the need for alternative strategies.

Rationalization Strategy

• Use conjugate to simplify limits: [ \frac{\sqrt{x^2 + 2x - 1} - \sqrt{x^2 - 1}}{1 + \sqrt{x^2 + 2x - 1} + \sqrt{x^2 - 1}} ]

• The simplified expression gives: [ \frac{2x}{\text{denominator}} ]

Simplification

• After simplification:

  • Rewrite numerator using ( \sqrt{x^2} ).

  • Since ( x ) is negative: ( |x| = -x )

  • Cancel ( x ) for final results: [ \frac{2}{-2} = -1 ]

Graphical Verification

• Graph confirms behavior: as ( x \to -\infty ), ( s \to -1 ) with horizontal asymptote at ( y = -1 ).

Trigonometric Limits

• Key identity:

  • ( \lim_{{x \to 0}} \frac{\sin x}{x} = 1 )

• Other useful limits:

  • ( \lim_{{x \to 0}} \frac{x}{\tan x} = 1 )

  • ( \lim_{{x \to 0}} \frac{1 - \cos x}{x} = 0 )

Example Problem: ( \frac{1 - \cos x}{x} )

• Direct substitution yields ( \frac{0}{0} ). Further analysis needed.

• Multiply and divide by ( 1 + \cos x ) to manipulate the expression: [ \frac{1 - \cos^2 x}{x (1 + \cos x)} = \frac{\sin^2 x}{x(1 + \cos x)} ]

• Rewrite as: [ \frac{\sin x}{x} \cdot \frac{\sin x}{x(1 + \cos x)} ]

• As ( x \to 0 ), this evaluates to:

  • Limit of ( \sin x / x ) becomes 1, thus:

  • Final result: 0 times a limit gives ultimately 0.

Trigonometric Practices

• Avoid incorrect assumptions about limits.

• Always check conditions under which sine and cosine functions operate.