Problem 285
Question
$$ \lim _{x \rightarrow \pm \infty} \cosh x-\sinh x\\{\text { Ans. } 0,+\infty\\} $$
Step-by-Step Solution
Verified Answer
Therefore, the limits of the function as \(x\) approaches positive infinity and negative infinity are \(0\) and \(+\infty\) respectively.
1Step 1: Understanding the given function
Firstly, we have the given function which is \(\cosh x - \sinh x\). We know that the definition of hyperbolic cosine \(\cosh x = 0.5(e^x + e^{-x})\) and hyperbolic sine \(\sinh x = 0.5(e^x - e^{-x})\). So, we substitute these equations into the given function, giving us \(0.5(e^x + e^{-x}) - 0.5(e^x - e^{-x})\). Simplify to find the function \(e^{-x}\).
2Step 2: Calculating the limit as \(x \rightarrow +\infty\)
To find the limit as \(x\) approaches positive infinity of \(e^{-x}\), we can just substitute \(+\infty\) for \(x\). Solving this yields \(e^{-\infty}\) which simplifies to \(0\).
3Step 3: Calculating the limit as \(x \rightarrow -\infty\)
Similarly, to find the limit as \(x\) approaches negative infinity, we substitute \(-\infty\) for \(x\), giving us \(e^{\infty}\), which simplifies to \(+\infty\) as per the properties of the exponential function.
Key Concepts
Hyperbolic FunctionsExponential FunctionsInfinity in Calculus
Hyperbolic Functions
Hyperbolic functions serve a similar role to trigonometric functions but are based on hyperbolas instead of circles. The main hyperbolic functions are the hyperbolic sine, written as \( \sinh x \), and hyperbolic cosine, written as \( \cosh x \). These functions are defined through the exponential function as follows:
\(\cosh x = \frac{e^x + e^{-x}}{2}\) and \( \sinh x = \frac{e^x - e^{-x}}{2} \). These formulas illustrate the symmetric relationship with their trigonometric analogs like sine and cosine, often used in calculus and other fields like physics.
In calculus, hyperbolic functions help to express complex behavior elegantly, simplifying many differential equations. They are intimately related to the exponential function, which allows for efficient computation of limits. For the given problem, we used the definitions of hyperbolic functions to transform \( \cosh x - \sinh x \) into a simpler form of \( e^{-x} \).
\(\cosh x = \frac{e^x + e^{-x}}{2}\) and \( \sinh x = \frac{e^x - e^{-x}}{2} \). These formulas illustrate the symmetric relationship with their trigonometric analogs like sine and cosine, often used in calculus and other fields like physics.
In calculus, hyperbolic functions help to express complex behavior elegantly, simplifying many differential equations. They are intimately related to the exponential function, which allows for efficient computation of limits. For the given problem, we used the definitions of hyperbolic functions to transform \( \cosh x - \sinh x \) into a simpler form of \( e^{-x} \).
Exponential Functions
Exponential functions are essential in both mathematics and applied sciences due to their unique properties. They are of the form: \( f(x) = e^x \), where \( e \) is the base of natural logarithms, approximately equal to 2.718.
This function is special due to its pervasive nature in growth and decay processes, like compound interest and population growth. The exponential function grows extraordinarily fast as \( x \) becomes large. Conversely, the inverse of the exponential function, \( e^{-x} \), decays rapidly.
In this exercise, we observe these characteristics. For \( x \rightarrow +\infty \), \( e^{-x} \) declines towards zero. Meanwhile, for \( x \rightarrow -\infty \), the function \( e^x \) grows without bound, illustrating how exponential functions approach infinity or zero based on the sign of the exponent.
This function is special due to its pervasive nature in growth and decay processes, like compound interest and population growth. The exponential function grows extraordinarily fast as \( x \) becomes large. Conversely, the inverse of the exponential function, \( e^{-x} \), decays rapidly.
In this exercise, we observe these characteristics. For \( x \rightarrow +\infty \), \( e^{-x} \) declines towards zero. Meanwhile, for \( x \rightarrow -\infty \), the function \( e^x \) grows without bound, illustrating how exponential functions approach infinity or zero based on the sign of the exponent.
Infinity in Calculus
Infinity in calculus is a concept used to describe bounds that go beyond any finite measurement. It is not a number but an idea that describes unbounded behavior in functions or sequences.
When evaluating limits, especially as \( x \rightarrow \infty \) or \( x \rightarrow -\infty \), we use this concept to understand the behavior of functions at extreme values. It allows mathematicians to predict the long-term trend of a function without calculating each value.
In the context of the exercise, \( e^{-x} \) approaches 0 as \( x \rightarrow +\infty \), reflecting how the function compresses towards the horizontal axis at infinity. Conversely, for \( x \rightarrow -\infty \), \( e^x \) surges towards infinity, demonstrating unbounded growth. These insights help solve limits and provide a clear comprehension of function behavior across different limit scenarios.
When evaluating limits, especially as \( x \rightarrow \infty \) or \( x \rightarrow -\infty \), we use this concept to understand the behavior of functions at extreme values. It allows mathematicians to predict the long-term trend of a function without calculating each value.
In the context of the exercise, \( e^{-x} \) approaches 0 as \( x \rightarrow +\infty \), reflecting how the function compresses towards the horizontal axis at infinity. Conversely, for \( x \rightarrow -\infty \), \( e^x \) surges towards infinity, demonstrating unbounded growth. These insights help solve limits and provide a clear comprehension of function behavior across different limit scenarios.
Other exercises in this chapter
Problem 283
$$ \lim _{x \rightarrow 0} \frac{e^{\sin 2 x}-e^{\sin x}}{x}\\{\text { Ans. } 1\\} $$
View solution Problem 284
$$ \lim _{x \rightarrow 0} \frac{\cos x-e^{-\frac{x^{2}}{2}}}{x^{4}}\left\\{\text { Ans. }-\frac{1}{12}\right\\} $$
View solution Problem 286
$$ \lim _{x \rightarrow \pm \infty} \tanh x \quad\\{\text { Ans. } 1,-1\\} $$
View solution Problem 287
$$ \lim _{x \rightarrow \pm \infty} x\left(\sqrt{x^{2}+\sqrt{x^{4}+1}}-x \sqrt{2}\right)\\{\text { Ans. } 0,-\infty\\} $$
View solution