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Question 01 (i) 

ShortAnswer Question. Show all your work, simplify your answer as much as possible.
Evaluate 
Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you? 
If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint. 
Hint 1 

At first one might think that L'Hopital's Rule would be a good strategy because evaluating the limit at zero gives an indeterminate form. However, taking the derivative of the denominator involves the chain rule, which will not help simplify the problem. Instead, try using a power series expansion of sinx in the numerator. 
Hint 2 

Once you have simplified the expression inside the limit using Hint 1, it is useful to recall the following known limit:
where can be replaced by any continuous function whenever . For example: . 
Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work.

Solution 

Recall the Maclaurin (power) series of sin(x) is
Note that the first two terms cancel exactly with the polynomial in the numerator. Thus, Now, as shown in Hint 2, we can also say that So the first term of the limit becomes . The remaining terms all have an x^{5} / sin(x^{5}) term which evaluates to 1, but also have an extra power of x which will make the entire term go to 0. So the solution to the limit is . This question demonstrates the strength of power series in more difficult calculus problems. Found a typo? Is this solution unclear? Let us know here, visit the discussion or suggest an alternative solution.

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