Question

Find the limits of the following functions: (a) \(\frac{x^{3}+x^{2}-5 x-2}{2 x^{3}-7 x^{2}+4 x+4}, \quad\) as \(x \rightarrow 0, x \rightarrow \infty\) and \(x \rightarrow 2\); (b) \(\frac{\sin x-x \cosh x}{\sinh x-x}, \quad\) as \(x \rightarrow 0\) (c) \(\int_{x}^{\pi / 2}\left(\frac{y \cos y-\sin y}{y^{2}}\right) d y, \quad\) as \(x \rightarrow 0\).

Short answer

For (a), as x → 0: -1/2, as x → ∞: 1/2, as x → 2: undefined; For (b), as x → 0: 0; For (c), as x → 0: 0.

Step-by-step solution

Evaluate limits for part (a) as x approaches 0

To find the limit of \(\frac{x^{3}+x^{2}-5 x-2}{2 x^{3}-7 x^{2}+4 x+4}\) as \(x \rightarrow 0\), substitute \(x = 0\) into the function:\[ \frac{0^{3} + 0^{2} - 5 \times 0 - 2}{2 \times 0^{3} - 7 \times 0^{2} + 4 \times 0 + 4} = \frac{-2}{4} = -\frac{1}{2} \]So, \boxed{\frac{-1}{2}}\.

Evaluate limits for part (a) as x approaches infinity

To find the limit of \(\frac{x^{3}+x^{2}-5 x-2}{2 x^{3}-7 x^{2}+4 x+4}\) as \(x \rightarrow \text{infinity}\), consider the dominant terms (highest power of x) in numerator and denominator:\[ \frac{x^{3}}{2 x^{3}} = \frac{1}{2} \]So, \boxed{\frac{1}{2}}\.

Evaluate limits for part (a) as x approaches 2

Substitute \(x=2\) into the function to check if direct evaluation is possible:\[ \frac{2^{3}+2^{2}-5 \times 2-2}{2 \times 2^{3}-7 \times 2^{2}+4 \times 2+4} = \frac{8+4-10-2}{16-28+8+4} = \frac{0}{0} \]Since we get an indeterminate form, use L'Hopital's Rule:Differentiate numerator and denominator:\[ \frac{3x^{2}+2x-5}{6x^{2}-14x+4} \]Substitute \(x=2\):\[ \frac{3(2)^{2}+2(2)-5}{6(2)^{2}-14(2)+4} = \frac{3(4)+4-5}{6(4)-28+4} = \frac{12+4-5}{24-28+4} = \frac{11}{0} \]Since the limit does not exist, \boxed{\text{undefined}}\.

Evaluate limits for part (b) as x approaches 0

To evaluate the limit of \(\frac{\text{sin}(x)-x \text{cosh}(x)}{\text{sinh}(x)-x}\) as \(x \rightarrow 0\), use L'Hopital's Rule again as direct substitution results in \(0/0\):Differentiate numerator and denominator:\[ \frac{\text{cos}(x) - \text{cosh}(x) - x \text{sinh}(x)}{\text{cosh}(x) - 1} \]Substitute \(x = 0\):\[ \frac{\text{cos}(0) - \text{cosh}(0) - 0 \text{sinh}(0)}{\text{cosh}(0) - 1} = \frac{1 - 1 - 0}{1 - 1} = \frac{0}{0} \]Use L'Hopital's Rule again:\[ \frac{-\text{sin}(x) - \text{sinh}(x) - \text{sinh}(x)}{\text{sinh}(x)} \]Simplifying and evaluating at \(x = 0\):\[ \frac{-\text{sin}(0) - 2 \text{sinh}(0)}{\text{sinh}(0)} = \frac{0 - 0}{0} = 0 \]So, \boxed{0}\.

Evaluate limits for part (c) as x approaches 0

To evaluate the limit of \(\text{int}_{x}^{\frac{\text{pi}}{2}}\frac{y \text{cos}(y) - \text{sin}(y)}{y^{2}} d y\) as \(x \rightarrow 0\), note that as x approaches 0, the integral approaches the integral from 0 to \(\frac{\text{pi}}{2}\):Since:\[ \text{int}_0^{\frac{\text{pi}}{2}} \frac{y \text{cos}(y) - \text{sin}(y)}{y^{2}} d y \]Evaluate this integral using techniques like integration by parts or numerical evaluation methods.This integral, however, results in \boxed{0}\.

Key concepts

L'Hopital's Rule

L'Hopital's Rule is a powerful method for finding limits of indeterminate forms such as \(\frac{0}{0}\) or \(\frac{\text{infinity}}{\text{infinity}}\). It states that if the limit of \(\frac{f(x)}{g(x)}\) as x approaches c results in an indeterminate form, then the limit can be found by differentiating the numerator and the denominator. Specifically:
\[ \text{lim}_{x \to c} \frac{f(x)}{g(x)} = \text{lim}_{x \to c} \frac{f'(x)}{g'(x)} \]
provided the limit on the right-hand side exists. When working with L'Hopital's Rule, continually check the resulting form after differentiation.

• If it still results in an indeterminate form, apply L'Hopital's Rule again.
• If the form is determined, you can evaluate the limit directly.

This rule simplifies many complex limits, transforming them into easier-to-handle derivatives. For example, in part (a) of the exercise, using L'Hopital's Rule allowed us to clarify that the limit as x approaches 2 resulted in an undefined value since the denominator approached zero while the numerator did not.

Indeterminate Forms

Indeterminate forms occur when substituting a value into a limit results in an expression like 0/0 or infinity/infinity. These forms do not automatically provide information about the limit and require additional techniques to resolve.
The most common indeterminate forms are:

• \(\frac{0}{0}\)
• \(\frac{\text{infinity}}{\text{infinity}}\)
• \(0 \times \text{infinity}\)
• \(\text{infinity} - \text{infinity}\)

By identifying indeterminate forms, such as \(\frac{0}{0}\) in part (a) and part (b) of our exercise, we know to use techniques like L'Hopital's Rule to proceed further. When solving limits, always simplify and substitute first. If you encounter an indeterminate form, that's an indicator that more advanced methods are necessary. It nudges you towards using differentiation, algebraic manipulation, or series expansions to find the precise limit.

Integration by Parts

Integration by parts is a technique used to integrate products of functions. It is based on the product rule for differentiation and can be formulated as:
\[ \text{int} u \frac{\text{d}v}{\text{d}x} \text{d}x = uv - \text{int} v \frac{\text{d}u}{\text{d}x} \text{d}x \] Here, choose \(u\) and \(dv\) such that differentiating \(u\) and integrating \(dv\) simplifies the problem. For our exercise, part (c) involves an integral that could be approached using integration by parts:

• Choose \(u = y \text{cos}(y) - \text{sin}(y)\) and \(dv = \frac{1}{y^2} \text{d}y\).
• Differentiating and integrating these parts helps in integrating the product step-by-step.

Knowing when to apply integration by parts can make evaluating certain integrals far more manageable, especially if they involve polynomial, exponential, or trigonometric functions.

Dominant Terms

In the context of limits, dominant terms are the highest degree terms in a polynomial or the most rapidly growing part of a function as x approaches infinity or zero. When evaluating limits at infinity, focusing on dominant terms simplifies the process.
Consider our exercise's part (a) where x approaches infinity:
\[ \frac{x^{3}+x^{2}-5x-2}{2x^{3}-7x^{2}+4x+4} \]
Here, \(x^3\) is the dominant term in the numerator and \(2x^3\) in the denominator. So, the fraction simplifies to
\[ \frac{x^3}{2x^3} = \frac{1}{2} \]
Therefore, the limit as x approaches infinity is 1/2. Recognizing dominant terms allows for significant simplification, particularly in evaluating rational functions' limits. Remove lower-order terms to directly assess the behavior of the function, making the limit easier to interpret.