lim x->0 sinbx/x
To evaluate the limit as x approaches 0 of sin(bx)/x, where b is a constant, we can use L’Hôpital’s Rule
To evaluate the limit as x approaches 0 of sin(bx)/x, where b is a constant, we can use L’Hôpital’s Rule. L’Hôpital’s Rule states that if the limit of a ratio of two functions is of the form 0/0 or ∞/∞, then taking the derivative of the numerator and denominator separately may help in determining the limit.
To apply L’Hôpital’s Rule to the given limit, let’s first write the limit using the standard limit notation:
lim x->0 sin(bx)/x
Now, we can rewrite sin(bx)/x as (1/x) * sin(bx). Applying L’Hôpital’s Rule involves taking the derivative of the numerator and denominator.
The derivative of (1/x) is given by:
d/dx(1/x) = -1/x^2
The derivative of sin(bx) is given by:
d/dx(sin(bx)) = b*cos(bx)
Now, let’s rewrite the limit using the derived derivatives:
lim x->0 (1/x) * sin(bx) = lim x->0 (1/x^2) * (b*cos(bx))
Taking the limit as x approaches 0 of (1/x^2) and (b*cos(bx)), we get:
lim x->0 (1/x^2) = ∞ (since as x approaches 0, the reciprocal of x^2 becomes larger and larger)
lim x->0 (b*cos(bx)) = b*cos(0) = b*1 = b (since the cosine of 0 is 1)
Now, multiplying the evaluated limits of (1/x^2) and (b*cos(bx)), we have:
lim x->0 (1/x^2) * (b*cos(bx)) = ∞ * b
Therefore, the limit as x approaches 0 of sin(bx)/x is ∞ * b, which simplifies to ∞ if b is a nonzero constant.
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