Improper Integral: Limit Can Be Zero Explained

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SANGHERA.JAS
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How the following limit can be zero, since after applying L'Hospital rule the root x will be in the numerator, which together with sinx will constitute infinity.
 

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I'm not sure you are actually using the product rule correctly. Hard to tell since you don't show what you actually did. But you have an x in the denominator. It's still infinity/infinity.
 
The proceedure is following, In my book they have equate it to zero.
 

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Your limit is completely wrong, the limit is 0. (1-cos(x))/x^1/2 goes to zero. Why do you think it doesn't? (1-cos(x)) is bounded. x^(1/2) isn't.
 
Oh, I got it. Thanks a lot.
 

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