QuestionJuly 26, 2025

lim _(xarrow 0^+)((3x+1)/(x)-(1)/(sinx))

lim _(xarrow 0^+)((3x+1)/(x)-(1)/(sinx))
lim _(xarrow 0^+)((3x+1)/(x)-(1)/(sinx))

Solution
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Answer

3 Explanation 1. Simplify the expression Combine the fractions: \frac{3x+1}{x} - \frac{1}{\sin x} = \frac{(3x+1)\sin x - x}{x \sin x}. 2. Apply L'Hôpital's Rule Since both numerator and denominator approach 0 as x \to 0^+, apply L'Hôpital's Rule: Differentiate numerator and denominator. 3. Differentiate the numerator Derivative of (3x+1)\sin x - x is 3\sin x + (3x+1)\cos x - 1. 4. Differentiate the denominator Derivative of x \sin x is \sin x + x \cos x. 5. Evaluate the limit Substitute x \to 0^+ into \frac{3\sin x + (3x+1)\cos x - 1}{\sin x + x \cos x} to get \frac{3(0) + (3(0)+1)(1) - 1}{0 + 0} = \frac{0}{0}, apply L'Hôpital's Rule again. 6. Apply L'Hôpital's Rule again Differentiate again: Numerator becomes 3\cos x - (3x+1)\sin x + 3\cos x, Denominator becomes \cos x + \cos x - x\sin x. 7. Evaluate the second derivative limit Substitute x \to 0^+ into \frac{6\cos x - (3x+1)\sin x}{2\cos x - x\sin x} to get \frac{6(1) - 0}{2(1) - 0} = \frac{6}{2} = 3.

Explanation

1. Simplify the expression<br /> Combine the fractions: $\frac{3x+1}{x} - \frac{1}{\sin x} = \frac{(3x+1)\sin x - x}{x \sin x}$.<br />2. Apply L'Hôpital's Rule<br /> Since both numerator and denominator approach 0 as $x \to 0^+$, apply L'Hôpital's Rule: Differentiate numerator and denominator.<br />3. Differentiate the numerator<br /> Derivative of $(3x+1)\sin x - x$ is $3\sin x + (3x+1)\cos x - 1$.<br />4. Differentiate the denominator<br /> Derivative of $x \sin x$ is $\sin x + x \cos x$.<br />5. Evaluate the limit<br /> Substitute $x \to 0^+$ into $\frac{3\sin x + (3x+1)\cos x - 1}{\sin x + x \cos x}$ to get $\frac{3(0) + (3(0)+1)(1) - 1}{0 + 0} = \frac{0}{0}$, apply L'Hôpital's Rule again.<br />6. Apply L'Hôpital's Rule again<br /> Differentiate again: Numerator becomes $3\cos x - (3x+1)\sin x + 3\cos x$, Denominator becomes $\cos x + \cos x - x\sin x$.<br />7. Evaluate the second derivative limit<br /> Substitute $x \to 0^+$ into $\frac{6\cos x - (3x+1)\sin x}{2\cos x - x\sin x}$ to get $\frac{6(1) - 0}{2(1) - 0} = \frac{6}{2} = 3$.
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