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Evaluate the limit.
lim_(x rarr0)(sen(8x)-x^(2)cos((6)/(x)))/(x)

Evaluate the limit. $\newline$ $\lim _{x \rightarrow 0} \frac{\operatorname{sin}(8 x)-x^{2} \cos \left(\frac{6}{x}\right)}{x}$

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Find derivatives of logarithmic functions

Full solution

Q. Evaluate the limit.

\newline

\lim _{x \rightarrow 0} \frac{\operatorname{sin}(8 x)-x^{2} \cos \left(\frac{6}{x}\right)}{x}

Identify Limit: Identify the limit that needs to be evaluated. $\newline$ We need to find the limit of the function $\frac{\sin(8x) - x^2 \cos(\frac{6}{x})}{x}$ as $x$ approaches $0$ .
Check Form: Check if the function is in an indeterminate form when $x$ approaches $0$ . Plugging in $x = 0$ , we get $(\sin(0) - 0^2 \cos(\frac{6}{0})) / 0$ , which is of the form $0/0$ , an indeterminate form.
Apply L'Hôpital's Rule: Apply L'Hôpital's Rule since we have an indeterminate form of $0/0$ . L'Hôpital's Rule states that if the limit of $f(x)/g(x)$ as $x$ approaches a value $c$ is of the form $0/0$ or $\infty/\infty$ , then the limit is the same as the limit of $f'(x)/g'(x)$ as $x$ approaches $c$ , provided that the latter limit exists.
Differentiate Numerator and Denominator: Differentiate the numerator and the denominator separately. $\newline$ The derivative of the numerator $\sin(8x)$ with respect to $x$ is $8\cos(8x)$ . $\newline$ The derivative of the numerator $-x^2 \cos(\frac{6}{x})$ with respect to $x$ is $-2x \cos(\frac{6}{x}) + (x^2)(\frac{6}{x^2})\sin(\frac{6}{x})$ which simplifies to $-2x \cos(\frac{6}{x}) + 6\sin(\frac{6}{x})$ . $\newline$ The derivative of the denominator $x$ with respect to $x$ is $1$ .
Apply Derivatives to Rule: Apply the derivatives to L'Hôpital's Rule. $\newline$ Now we need to evaluate the limit of $\frac{8\cos(8x) - 2x \cos(\frac{6}{x}) + 6\sin(\frac{6}{x})}{1}$ as $x$ approaches $0$ .
Evaluate Simplified Expression: Evaluate the limit of the simplified expression as $x$ approaches $0$ . As $x$ approaches $0$ , $\cos(8x)$ approaches $\cos(0)$ which is $1$ , and $\sin(\frac{6}{x})$ approaches $\sin(\infty)$ which oscillates and does not have a limit. However, since $\sin(\frac{6}{x})$ is multiplied by $x$ in the term $0$ $1$ , this term approaches $0$ . Therefore, the limit of the expression is $0$ $3$ , which simplifies to $0$ $4$ .

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