Limiting a Complex Function to π

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SUMMARY

The discussion focuses on evaluating the limit of the complex function as x approaches π, specifically the expression \lim_{\substack{x\rightarrow\pi}} {\left( \frac {x}{x-\pi}{\int_{\pi}^{x} }\frac{sin t}{t}} dt\right)}. The primary challenge arises from the indeterminate form "0/0" when applying the limit. Participants recommend using L'Hôpital's rule to resolve this, leading to the conclusion that the limit evaluates to the integrand at π, which is sin(π)/π = 0.

PREREQUISITES
  • Understanding of limits and continuity in calculus
  • Familiarity with L'Hôpital's rule for resolving indeterminate forms
  • Knowledge of integral calculus, specifically the Fundamental Theorem of Calculus
  • Basic understanding of trigonometric functions and their properties
NEXT STEPS
  • Study L'Hôpital's rule in-depth, including its applications and limitations
  • Explore the Fundamental Theorem of Calculus and its implications for evaluating limits
  • Investigate the behavior of trigonometric functions near their critical points
  • Practice solving various indeterminate forms using different calculus techniques
USEFUL FOR

Students and professionals in mathematics, particularly those studying calculus, as well as educators seeking to enhance their understanding of limit evaluation techniques.

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[tex]{{\lim_{\substack{x\rightarrow\pi}} {\left( \frac {x}{x-\pi}{\int_{\pi}^{x} }\frac{sin t}{t}} dt\right)}[/tex]
 
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First, of course, that "x" outside the integral goes to [itex]\pi[/itex]. The only problem is
[tex]\lim_{x\to\pi}\frac{\int_\pi^x \frac{sin(t)}{t} dt}{x- \pi}[/tex]
which gives the "0/0" indeterminate form.

Use L'Hopital's rule to find that limit.
 
HallsofIvy said:
First, of course, that "x" outside the integral goes to [itex]\pi[/itex]. The only problem is
[tex]\lim_{x\to\pi}\frac{\int_\pi^x \frac{sin(t)}{t} dt}{x- \pi}[/tex]
which gives the "0/0" indeterminate form.

Use L'Hopital's rule to find that limit.
The expression is simply the derivative of the integral, i.e. the integrand at π, which is sin(π)/π = 0.
 

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