Asked by Anthony
lim x -> infinity for (sin^2x)/(x^2-1)
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Answered by
Jai
limit of (sin^2 (x))/(x^2 - 1) as x -> infinity
Note that the range of values of sin x is from -1 to 1, for all real values of x. Thus the range of sin^2 (x) is only from 0 to 1 (as any real number squared results to greater than or equal to zero).
Now, for the denominator, as x -> infinity, x^2 approaches infinity also, and so a small number (between 0 and 1) divided by the denominator infinity is equal to zero:
lim (sin^2 (x))/(x^2 - 1) as x->infinity
= 0.
hope this helps- :3
Note that the range of values of sin x is from -1 to 1, for all real values of x. Thus the range of sin^2 (x) is only from 0 to 1 (as any real number squared results to greater than or equal to zero).
Now, for the denominator, as x -> infinity, x^2 approaches infinity also, and so a small number (between 0 and 1) divided by the denominator infinity is equal to zero:
lim (sin^2 (x))/(x^2 - 1) as x->infinity
= 0.
hope this helps- :3
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