watchmath
  • watchmath
Compute \[\lim_{x\to 2}\frac{\int_{-2}^x \sin(t^3)\,dt}{x^4-16}\]
Mathematics
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watchmath
  • watchmath
Compute \[\lim_{x\to 2}\frac{\int_{-2}^x \sin(t^3)\,dt}{x^4-16}\]
Mathematics
schrodinger
  • schrodinger
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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anonymous
  • anonymous
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
watchmath
  • watchmath
Great! It is not very clear from above that what I wrote was sin(x^3)
anonymous
  • anonymous
oh allright then sin(8)/32

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