Answer:
π
Explanation:
Solving without L'Hopital's rule:
lim(x→0) sin(π cos²x) / x²
Use Pythagorean identity:
lim(x→0) sin(π (1 − sin²x)) / x²
lim(x→0) sin(π − π sin²x) / x²
Use angle difference formula:
lim(x→0) [ sin(π) cos(-π sin²x) − cos(π) sin(-π sin²x) ] / x²
lim(x→0) -sin(-π sin²x) / x²
Use angle reflection formula:
lim(x→0) sin(π sin²x) / x²
Now we multiply by π sin²x / π sin²x.
lim(x→0) [ sin(π sin²x) / x² ] (π sin²x / π sin²x)
lim(x→0) [ sin(π sin²x) / π sin²x] (π sin²x / x²)
lim(x→0) [ sin(π sin²x) / π sin²x] lim(x→0) (π sin²x / x²)
π lim(x→0) [ sin(π sin²x) / π sin²x] [lim(x→0) (sin x / x)]²
Use identity lim(u→0) (sin u / u) = 1.
π (1) (1)²
π
Solving with L'Hopital's rule:
If we plug in x = 0, the limit evaluates to 0/0. So using L'Hopital's rule:
lim(x→0) [ cos(π cos²x) (-2π cos x sin x) ] / 2x
lim(x→0) [ -π cos(π cos²x) sin(2x) ] / 2x
-π/2 lim(x→0) [ cos(π cos²x) sin(2x) ] / x
Again, the limit evaluates to 0/0. So using L'Hopital's rule one more time:
-π/2 lim(x→0) [ cos(π cos²x) (2 cos(2x)) + (-sin(π cos²x) (-2π cos x sin x)) sin(2x) ] / 1
-π/2 lim(x→0) [ 2 cos(π cos²x) cos(2x) + π sin(π cos²x) sin²(2x) ]
-π/2 (-2)
π