Answer: -1
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Step-by-step explanation:
I believe the given limit is
![\displaystyle \lim_{x\to\infty} \bigg(\sqrt[3]{3x^3+3x^2+x-1} - \sqrt[3]{3x^3-x^2+1}\bigg)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Clim_%7Bx%5Cto%5Cinfty%7D%20%5Cbigg%28%5Csqrt%5B3%5D%7B3x%5E3%2B3x%5E2%2Bx-1%7D%20-%20%5Csqrt%5B3%5D%7B3x%5E3-x%5E2%2B1%7D%5Cbigg%29)
Let

Now rewrite the expression as a difference of cubes:

Then

The limit is then equivalent to

From each remaining cube root expression, remove the cubic terms:



Now that we see each term in the denominator has a factor of <em>x</em> ², we can eliminate it :


As <em>x</em> goes to infinity, each of the 1/<em>x</em> ⁿ terms converge to 0, leaving us with the overall limit,

Answer:
28
Step-by-step explanation:
Answer:
d≈2yd
Step-by-step explanation:
Answer:
The running time is quadratic (O(n²) )
Step-by-step explanation:
For the set up, we have a constant running time of C. The, a log-linearsorting is called, thus, its execution time, denoted by T(n), is O(n*log(n)). Then, we call n times a linear iteration, with a running time of an+b, for certain constants a and b, thus, the running time of the algorithm is
C + T(n) + n*(a*n+b) = an²+bn + T + C
Since T(n) is O(n*log(n)) and n² is asymptotically bigger than n*log(n), then the running time of the algorith is quadratic, therefore, it is O(n²).