1.3 Evaluating limits analytically; Question 100

[Eigenvalue]

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[Eigenvalue]

As substituting x=0 evaluates to the indeterminate form [tex]\frac{0}{0}[/tex], L'hoptital's rule can be used to determine the limit of the expression. [tex]\lim_{x \to 0} \frac{sin²(x)}{x}[/tex] can be rewritten as [tex]\lim_{x \to 0}2sin(x)cos(x)[/tex], given that the derivative of x is 1, and the derivative of sin²(x) is 2sin(x)cos(x), due to the chain rule. 2sin(x)cos(x) evaluates to sin(2x), therefore, the expression can be further simplified to [tex]\lim_{x \to 0}sin(2x)[/tex]. Substituting x=0 for [tex]\lim_{x \to 0} \frac{sin²(x)}{x}[/tex] evaluates to sin(0), or 0.

[nycmath]

As substituting x=0 evaluates to the indeterminate form [tex]\frac{0}{0}[/tex], L'hoptital's rule can be used to determine the limit of the expression. [tex]\lim_{x \to 0} \frac{sin²(x)}{x}[/tex] can be rewritten as [tex]\lim_{x \to 0}2sin(x)cos(x)[/tex], given that the derivative of x is 1, and the derivative of sin²(x) is 2sin(x)cos(x), due to the chain rule. 2sin(x)cos(x) evaluates to sin(2x), therefore, the expression can be further simplified to [tex]\lim_{x \to 0}sin(2x)[/tex]. Substituting x=0 for [tex]\lim_{x \to 0} \frac{sin²(x)}{x}[/tex] evaluates to sin(0), or 0.

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