asked 168k views
5 votes
Evaluate the limit lim(x → 0) tan(πx) ln(1/x)

A) 0
B) -π/2
C) [infinity]
D) Does not exist

asked
User MaTHwoG
by
7.9k points

1 Answer

4 votes

Final answer:

The limit of tan(πx) ln(1/x) as x approaches 0 is evaluated using L'Hôpital's rule, resulting in the answer A) 0 because as x approaches 0, tan(πx) approaches 0 and ln(1/x) approaches infinity, forming an indeterminate form that is resolved to 0.

Step-by-step explanation:

The question asks to evaluate the limit of tan(πx) ln(1/x) as x approaches 0. We can use the property that π is a constant and the fact that tangent function is odd, hence tan(πx) approaches 0 as x → 0. However, the natural logarithm ln(1/x) approaches infinity as x → 0 because as x gets smaller, 1/x gets larger without bound. Therefore, we have a 0 times infinity situation, which is an indeterminate form. To resolve this, we can use L'Hôpital's rule which transforms the original limit into the limit of the derivatives of the top and bottom functions:

  • Derivative of tan(πx) with respect to x is πsec2(πx).
  • Derivative of ln(1/x) with respect to 1/x is -1/x2.

By applying L'Hôpital's rule, the new limit is:

limx → 0 (πsec2(πx) / (-1/x2)) which simplifies to limx → 0 (-πx2sec2(πx)). Since sec(πx) approaches 1 and x2 approaches 0 as x approaches 0, the resulting limit is 0. Thus, the correct answer is A) 0.

answered
User Julius Delfino
by
8.2k points
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