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lim_(x rarr0^(+))(9x)^(sin(3x))

lim⁡x→0+(9x)sin⁡(3x) \lim _{x \rightarrow 0^{+}}(9 x)^{\sin (3 x)}

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Full solution

Q. lim⁡x→0+(9x)sin⁡(3x) \lim _{x \rightarrow 0^{+}}(9 x)^{\sin (3 x)}
  1. Rewrite Limit Property: Rewrite the limit using the property that eln⁡(a)=ae^{\ln(a)} = a.lim⁡x→0+(9x)sin⁡(3x)=lim⁡x→0+eln⁡((9x)sin⁡(3x))\lim_{x \rightarrow 0^{+}}(9x)^{\sin(3x)} = \lim_{x \rightarrow 0^{+}}e^{\ln((9x)^{\sin(3x)})}
  2. Logarithmic Property Application: Use the property of logarithms that ln⁡(ab)=b⋅ln⁡(a)\ln(a^b) = b\cdot\ln(a).eln⁡((9x)sin⁡(3x))=esin⁡(3x)⋅ln⁡(9x)e^{\ln((9x)^{\sin(3x)})} = e^{\sin(3x)\cdot\ln(9x)}
  3. Focus on Limit Analysis: Since the limit of sin⁡(3x)\sin(3x) as xx approaches 00 is 00, we can focus on the limit of ln⁡(9x)\ln(9x) as xx approaches 00.
    lim⁡x→0+sin⁡(3x)⋅ln⁡(9x)=lim⁡x→0+sin⁡(3x)⋅lim⁡x→0+ln⁡(9x)\lim_{x \to 0^{+}}\sin(3x)\cdot\ln(9x) = \lim_{x \to 0^{+}}\sin(3x) \cdot \lim_{x \to 0^{+}}\ln(9x)
  4. Solve Indeterminate Form: The limit of sin⁡(3x)\sin(3x) as xx approaches 00 is 00.lim⁡x→0+sin⁡(3x)=0\lim_{x \rightarrow 0^{+}}\sin(3x) = 0
  5. Apply L'Hôpital's Rule: The limit of ln⁡(9x)\ln(9x) as xx approaches 00 from the positive side is negative infinity.\newlinelim⁡x→0+ln⁡(9x)=−∞\lim_{x \to 0^{+}}\ln(9x) = -\infty
  6. Differentiate Numerator and Denominator: The product of 00 and negative infinity is an indeterminate form, so we need to use L'Hôpital's Rule.\newlineWe differentiate the numerator and denominator of sin⁡(3x)1/x.\frac{\sin(3x)}{1/x}.
  7. Simplify Expression: Differentiate sin⁡(3x)\sin(3x) to get 3cos⁡(3x)3\cos(3x) and differentiate 1x\frac{1}{x} to get −1x2-\frac{1}{x^2}. \newlinelim⁡x→0+(3cos⁡(3x)−1x2)\lim_{x \to 0^{+}}\left(\frac{3\cos(3x)}{-\frac{1}{x^2}}\right)
  8. Evaluate Limit Result: Simplify the expression by multiplying both numerator and denominator by x2x^2. \newlinelim⁡x→0+(3x2cos⁡(3x)−1)\lim_{x \rightarrow 0^{+}}\left(\frac{3x^2\cos(3x)}{-1}\right)
  9. Evaluate Limit Result: Simplify the expression by multiplying both numerator and denominator by x2x^2. \newlinelim⁡x→0+(3x2cos⁡(3x)−1)\lim_{x \rightarrow 0^{+}}\left(\frac{3x^2\cos(3x)}{-1}\right)The limit of 3x2cos⁡(3x)3x^2\cos(3x) as xx approaches 00 is 00.\newlinelim⁡x→0+(3x2cos⁡(3x))=0\lim_{x \rightarrow 0^{+}}(3x^2\cos(3x)) = 0
  10. Evaluate Limit Result: Simplify the expression by multiplying both numerator and denominator by x2x^2. \newlinelim⁡x→0+(3x2cos⁡(3x)−1)\lim_{x \to 0^{+}}\left(\frac{3x^2\cos(3x)}{-1}\right)The limit of 3x2cos⁡(3x)3x^2\cos(3x) as xx approaches 00 is 00. \newlinelim⁡x→0+(3x2cos⁡(3x))=0\lim_{x \to 0^{+}}(3x^2\cos(3x)) = 0The limit of the entire expression is 0(−1)\frac{0}{(-1)}, which is 00. \newlinelim⁡x→0+(9x)sin⁡(3x)=0\lim_{x \to 0^{+}}(9x)^{\sin(3x)} = 0

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