Question

Hiiii.. please help me with this limit question ​

Answer 1

Answer:

π

Step-by-step 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