Question

lim x=>0 sin(π cos²x)/x²​

Answer 1

Answer:

Since the question is in form of (0/0)

Applying l hospital rule....

So, differentiating it as

Lim x--0 [{cos(πcos^x) π(-sin2x)}/2x]

Now, Lim x--0 (-π) cos(πcos^x)*(sin2x)/2x

Hence , (-π) *(-1)*1=π

Ans=π.

Step-by-step explanation:

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