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watchmath

  • 5 years ago

Compute \[\lim_{x\to 2}\frac{\int_{-2}^x \sin(t^3)\,dt}{x^4-16}\]

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  1. anonymous
    • 5 years ago
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    sin is odd, so as x goes to 2 the top and bottom are both going to 0. We can use L'hopitals rule and differentiate the numerator and denominator. On top we get sin(x^2). on the bottom we get 4x^3. Now we can plug in x=2. the limit as x goes to 2 is sin(4)/(32). Let me know if what I did is ok

  2. watchmath
    • 5 years ago
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    Great! It is not very clear from above that what I wrote was sin(x^3)

  3. anonymous
    • 5 years ago
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    oh allright then sin(8)/32

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