a. $\lim _{n \rightarrow 0} \frac{\sin x-x}{x^{3}}$ $\newline$ b. $\int_{0}^{1 / 2} \frac{\ln (x+1)}{x} d x=$

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Evaluate definite integrals using the power rule

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Q. a.

\lim _{n \rightarrow 0} \frac{\sin x-x}{x^{3}}

\newline

\int_{0}^{1 / 2} \frac{\ln (x+1)}{x} d x=

Apply L'Hôpital's Rule: To solve part a, we need to find the limit of $(\sin x - x) / x^3$ as $x$ approaches $0$ . We can use L'Hôpital's rule since the limit is of the indeterminate form $0/0$ .
Evaluate Limit: Apply L'Hôpital's rule once by taking the derivative of the numerator and the derivative of the denominator. $\newline$ The derivative of the numerator $\sin x - x$ is $\cos x - 1$ . $\newline$ The derivative of the denominator $x^3$ is $3x^2$ . $\newline$ Now the limit becomes $\lim_{x \to 0} \frac{\cos x - 1}{3x^2}$ .
Deal with Improper Integral: Apply L'Hôpital's rule a second time because the new limit is still of the indeterminate form $0/0$ . The derivative of the numerator $(\cos x - 1)$ is $-\sin x$ . The derivative of the denominator $3x^2$ is $6x$ . Now the limit becomes $\lim_{x \to 0} \frac{-\sin x}{6x}$ .
Set Up Limit: Apply L'Hôpital's rule a third time because the limit is still of the indeterminate form $0/0$ . The derivative of the numerator $-\sin x$ is $-\cos x$ . The derivative of the denominator $6x$ is $6$ . Now the limit becomes $\lim_{x \to 0} \frac{-\cos x}{6}$ .
Use Substitution: Evaluate the limit $\lim_{x \to 0} \frac{-\cos x}{6}$ . As $x$ approaches $0$ , $\cos x$ approaches $1$ , so the limit is $-\frac{1}{6}$ .
Use Substitution: Evaluate the limit $\lim_{x \to 0} \frac{-\cos x}{6}$ . As $x$ approaches $0$ , $\cos x$ approaches $1$ , so the limit is $-\frac{1}{6}$ .To solve part b, we need to evaluate the integral of $\frac{\ln(x+1)}{x}$ from $0$ to $\frac{1}{2}$ . This is an improper integral because $\ln(x+1)$ is undefined at $x$ $0$ and the integrand has a vertical asymptote at $x$ $1$ .
Use Substitution: Evaluate the limit $\lim_{x \to 0} \frac{-\cos x}{6}$ . As $x$ approaches $0$ , $\cos x$ approaches $1$ , so the limit is $-\frac{1}{6}$ .To solve part b, we need to evaluate the integral of $\frac{\ln(x+1)}{x}$ from $0$ to $\frac{1}{2}$ . This is an improper integral because $\ln(x+1)$ is undefined at $x$ $0$ and the integrand has a vertical asymptote at $x$ $1$ .To deal with the improper integral, we will take the limit as $x$ $2$ approaches $0$ from the right of the integral from $x$ $2$ to $\frac{1}{2}$ of $\frac{\ln(x+1)}{x}$ dx.
Use Substitution: Evaluate the limit $\lim_{x \to 0} \frac{-\cos x}{6}$ . As $x$ approaches $0$ , $\cos x$ approaches $1$ , so the limit is $-\frac{1}{6}$ .To solve part b, we need to evaluate the integral of $\frac{\ln(x+1)}{x}$ from $0$ to $\frac{1}{2}$ . This is an improper integral because $\ln(x+1)$ is undefined at $x$ $0$ and the integrand has a vertical asymptote at $x$ $1$ .To deal with the improper integral, we will take the limit as $x$ $2$ approaches $0$ from the right of the integral from $x$ $2$ to $\frac{1}{2}$ of $x$ $6$ .Set up the limit: $x$ $7$ .
Use Substitution: Evaluate the limit $\lim_{x \to 0} \frac{-\cos x}{6}$ . As $x$ approaches $0$ , $\cos x$ approaches $1$ , so the limit is $-\frac{1}{6}$ .To solve part b, we need to evaluate the integral of $\frac{\ln(x+1)}{x}$ from $0$ to $\frac{1}{2}$ . This is an improper integral because $\ln(x+1)$ is undefined at $x$ $0$ and the integrand has a vertical asymptote at $x$ $1$ .To deal with the improper integral, we will take the limit as $x$ $2$ approaches $0$ from the right of the integral from $x$ $2$ to $\frac{1}{2}$ of $x$ $6$ .Set up the limit: $x$ $7$ .To evaluate the integral, we recognize that it is a standard integral that can be solved using substitution. Let $x$ $8$ , which implies $x$ $9$ .
Use Substitution: Evaluate the limit $\lim_{x \to 0} \frac{-\cos x}{6}$ . As $x$ approaches $0$ , $\cos x$ approaches $1$ , so the limit is $-\frac{1}{6}$ .To solve part b, we need to evaluate the integral of $\frac{\ln(x+1)}{x}$ from $0$ to $\frac{1}{2}$ . This is an improper integral because $\ln(x+1)$ is undefined at $x$ $0$ and the integrand has a vertical asymptote at $x$ $1$ .To deal with the improper integral, we will take the limit as $x$ $2$ approaches $0$ from the right of the integral from $x$ $2$ to $\frac{1}{2}$ of $\frac{\ln(x+1)}{x}$ dx.Set up the limit: $x$ $7$ .To evaluate the integral, we recognize that it is a standard integral that can be solved using substitution. Let $x$ $8$ , which implies $x$ $9$ .Change the limits of integration according to the substitution. When $0$ $0$ , $0$ $1$ . When $0$ $2$ , $0$ $3$ .
Use Substitution: Evaluate the limit $\lim_{x \to 0} \frac{-\cos x}{6}$ . As $x$ approaches $0$ , $\cos x$ approaches $1$ , so the limit is $-\frac{1}{6}$ .To solve part b, we need to evaluate the integral of $\frac{\ln(x+1)}{x}$ from $0$ to $\frac{1}{2}$ . This is an improper integral because $\ln(x+1)$ is undefined at $x$ $0$ and the integrand has a vertical asymptote at $x$ $1$ .To deal with the improper integral, we will take the limit as $x$ $2$ approaches $0$ from the right of the integral from $x$ $2$ to $\frac{1}{2}$ of $\frac{\ln(x+1)}{x}$ dx.Set up the limit: $x$ $7$ .To evaluate the integral, we recognize that it is a standard integral that can be solved using substitution. Let $x$ $8$ , which implies $x$ $9$ .Change the limits of integration according to the substitution. When $0$ $0$ , $0$ $1$ . When $0$ $2$ , $0$ $3$ .The integral becomes $0$ $4$ .
Use Substitution: Evaluate the limit $\lim_{x \to 0} \frac{-\cos x}{6}$ . As $x$ approaches $0$ , $\cos x$ approaches $1$ , so the limit is $-\frac{1}{6}$ .To solve part b, we need to evaluate the integral of $\frac{\ln(x+1)}{x}$ from $0$ to $\frac{1}{2}$ . This is an improper integral because $\ln(x+1)$ is undefined at $x$ $0$ and the integrand has a vertical asymptote at $x$ $1$ .To deal with the improper integral, we will take the limit as $x$ $2$ approaches $0$ from the right of the integral from $x$ $2$ to $\frac{1}{2}$ of $x$ $6$ .Set up the limit: $x$ $7$ .To evaluate the integral, we recognize that it is a standard integral that can be solved using substitution. Let $x$ $8$ , which implies $x$ $9$ .Change the limits of integration according to the substitution. When $0$ $0$ , $0$ $1$ . When $0$ $2$ , $0$ $3$ .The integral becomes $0$ $4$ .This integral does not have a standard antiderivative and cannot be evaluated using elementary functions. It is known as the logarithmic integral and is typically expressed in terms of the special function $0$ $5$ .
Use Substitution: Evaluate the limit $\lim_{x \to 0} \frac{-\cos x}{6}$ . As $x$ approaches $0$ , $\cos x$ approaches $1$ , so the limit is $-\frac{1}{6}$ .To solve part b, we need to evaluate the integral of $\frac{\ln(x+1)}{x}$ from $0$ to $\frac{1}{2}$ . This is an improper integral because $\ln(x+1)$ is undefined at $x$ $0$ and the integrand has a vertical asymptote at $x$ $1$ .To deal with the improper integral, we will take the limit as $x$ $2$ approaches $0$ from the right of the integral from $x$ $2$ to $\frac{1}{2}$ of $x$ $6$ .Set up the limit: $x$ $7$ .To evaluate the integral, we recognize that it is a standard integral that can be solved using substitution. Let $x$ $8$ , which implies $x$ $9$ .Change the limits of integration according to the substitution. When $0$ $0$ , $0$ $1$ . When $0$ $2$ , $0$ $3$ .The integral becomes $0$ $4$ .This integral does not have a standard antiderivative and cannot be evaluated using elementary functions. It is known as the logarithmic integral and is typically expressed in terms of the special function $0$ $5$ .Since we cannot find an elementary antiderivative, we cannot proceed further with the evaluation of the integral. This is a math error because the problem requires an exact, simplified answer, which we cannot provide using elementary functions.