Evaluate the limit.
lim x → 1 ln x/sin 5πx
To evaluate the limit, we can try to apply direct substitution. However, we encounter an indeterminate form (0/0) when substituting x = 1 into the expression ln x / sin (5πx). Therefore, we need to use some algebraic manipulation and trigonometric identities to simplify the expression and evaluate the limit.
First, let's simplify the expression by using properties of logarithms:
lim x → 1 ln x / sin (5πx) = ln(1) / sin (5π(1))
Since ln(1) = 0, the numerator becomes zero.
Next, let's address the denominator sin (5π(1)):
sin (5π) = sin (π) = 0
Therefore, the denominator also evaluates to zero.
Now, we have the expression 0 / 0, which is still an indeterminate form. To further simplify and evaluate the limit, we can use L'Hopital's Rule.
Taking the derivative of the numerator and denominator separately, we get:
lim x → 1 ln x / sin (5πx) = lim x → 1 (1/x) / (5π cos (5πx))
Applying L'Hopital's Rule again, we differentiate numerator and denominator once more:
lim x → 1 (1/x) / (5π cos (5πx)) = lim x → 1 (-1/x^2) / (-5π^2 sin (5πx))
Simplifying further, we obtain:
lim x → 1 (-1/x^2) / (-5π^2 sin (5πx)) = lim x → 1 1/(5π^2 x^2 sin (5πx))
Since x → 1, we can use the property sin (5πx) ≈ 5πx - 5πx^3/3 for small values of x. Substituting this approximation into our expression, we have:
lim x → 1 1/(5π^2 x^2 sin (5πx)) ≈ lim x → 1 1/(5π^2 x^2 (5πx - 5πx^3/3))
Now, we can simplify the expression further by canceling out common factors:
lim x → 1 1/(5π^2 x^2 (5πx - 5πx^3/3)) = lim x → 1 1/(25π^3 x^3 - 25π^3 x^5/3)
Now, we can substitute x = 1 into the expression:
lim x → 1 1/(25π^3 x^3 - (25/3)π^3 x^5) = 1/(25π^3 - (25/3)π^3)
Simplifying further, we obtain:
lim x → 1 1/(25π^3 - (25/3)π^3) = 1/(75π^3 - 75π^3/3)
This simplifies to:
lim x → 1 1/(75π^3 - 25π^3) = 1/(50π^3)
Thus, the limit of ln x / sin (5πx) as x approaches 1 is 1/(50π^3).
To evaluate the limit, we can try to directly substitute the value of x into the expression. Let's plug in x = 1:
lim x → 1 ln x/sin 5πx = ln(1) / sin(5π(1))
Now, ln(1) is equal to 0 since the natural logarithm of 1 is always 0. And since sin(5π) is also 0, we have:
lim x → 1 ln x/sin 5πx = 0 / 0
We have encountered an indeterminate form of limit, which means we cannot determine the value of the limit just by plugging in the value of x.
To further evaluate this limit, we can use L'Hôpital's Rule. This rule states that if we have a limit of the form 0/0 or ∞/∞, we can take the derivative of the numerator and the derivative of the denominator, and then evaluate the limit again.
In this case, let's take the derivatives of ln x and sin 5πx with respect to x:
d/dx (ln x) = 1/x
d/dx (sin 5πx) = 5π cos 5πx
Now, we can rewrite the limit:
lim x → 1 ln x/sin 5πx = lim x → 1 (1/x)/(5π cos 5πx)
Let's plug in x = 1 now:
lim x → 1 (1/x)/(5π cos 5πx) = (1/1)/(5π cos 5π(1))
= 1/(5π cos 5π)
= 1/(5π * (-1))
= -1/(5π)
Therefore, the limit of ln x/sin 5πx as x approaches 1 is -1/(5π).
lim [ ln (x)/(sin (5πx)) ] as x → 1
Use L'hopital's rule. Differentiate numerator and denominator separately:
lim [ (1/x) / 5π(cos(5πx)) ] as x → 1
Substitute x = 1:
= (1/1) / 5π(cos(5π(1)))
= 1 / 5π(-1)
= -1 / 5π