What is the limit of tanh x as x approaches infinity?

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Homework Help Overview

The discussion revolves around evaluating the limit of the hyperbolic tangent function, tanh x, as x approaches infinity. Participants are exploring the mathematical properties and behaviors of the function using its definitions in terms of sinh and cosh.

Discussion Character

  • Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants attempt to evaluate the limit using the definitions of sinh and cosh, and some mention applying l'Hospital's rule. Others suggest alternative manipulations of the expression to simplify the limit evaluation.

Discussion Status

The discussion includes various approaches to the limit, with some participants providing different methods for simplification. There is no explicit consensus on the final outcome, but several lines of reasoning are being explored.

Contextual Notes

Some participants question the appropriateness of methods used, such as l'Hospital's rule, while others propose alternative algebraic manipulations. The nature of the problem suggests a focus on understanding the behavior of the function as x increases without bound.

merced
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\lim_{x\rightarrow\infty} tanh x = \lim_{x\rightarrow\infty} \frac{sinh x}{cosh x}
= \lim_{x\rightarrow\infty} \frac{e^x - e^{-x}}{e^x + e^{-x}}
= \lim_{x\rightarrow\infty} \frac{e^{2x} -1}{e^{2x} +1}

what now?
 
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Evaluate the limit.
 
merced said:
\lim_{x\rightarrow\infty} tanh x = \lim_{x\rightarrow\infty} \frac{sinh x}{cosh x}
= \lim_{x\rightarrow\infty} \frac{e^x - e^{-x}}{e^x + e^{-x}}
= \lim_{x\rightarrow\infty} \frac{e^{2x} -1}{e^{2x} +1}

what now?

=^{H} \lim_{x\rightarrow\infty} \frac{2e^{2x}}{2e^{2x}}=1

where the symbol =^{H} denotes the use of l'Hospital's rule.
 
thank you.
 
Swatting a fly with a sledgehammer!
Multiply both numerator and denominator of
= \lim_{x\rightarrow\infty} \frac{e^x - e^{-x}}{e^x + e^{-x}}
by e-x rather than ex and you get
= \lim_{x\rightarrow\infty} \frac{1- e^{-2x}}{1+ e^{-2x}}
which is obvious.
 

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