Question

Evaluate limit of hyperbolic cos by direct substitution ${{\lim }_{x\to 0}}(\cosh \,x)$.

Answer 1

Hint: We will substitute the formula of cosh x that is $f(x)=\dfrac{{{e}^{x}}+{{e}^{-x}}}{2}$ directly in ${{\lim }_{x\to 0}}(\cosh \,x)$ and then we will apply the limit and will get the answer.

Complete step-by-step answer:
Before proceeding with the above question we should understand about hyperbolic functions.
The hyperbolic functions are analogs of the circular function or the trigonometric functions. The hyperbolic function occurs in the solutions of linear differential equations, calculation of distance and angles in the hyperbolic geometry, Laplace’s equations in the cartesian coordinates. Generally, the hyperbolic function takes place in the real argument called the hyperbolic angle. The basic hyperbolic functions are: Hyperbolic sine (sinh), Hyperbolic cosine (cosh) and Hyperbolic tangent (tanh).
The hyperbolic cosine function is a function $f:R\to R$ is defined by $f(x)=\dfrac{{{e}^{x}}+{{e}^{-x}}}{2}$ and it is denoted by cosh x.
$\Rightarrow {{\lim }_{x\to 0}}(\cosh \,x)......(1)$
We know cosh x is equal to $\dfrac{{{e}^{x}}+{{e}^{-x}}}{2}$ , so substituting this in equation (1) we get,
$\Rightarrow {{\lim }_{x\to 0}}\dfrac{{{e}^{x}}+{{e}^{-x}}}{2}......(2)$
Now putting x equal to 0 in equation (2) we get,
$\Rightarrow {{\lim }_{x\to 0}}\dfrac{{{e}^{0}}+{{e}^{-0}}}{2}......(3)$
Now we know that any number having zero as power have value 1, so substituting this in equation (3) we get,
$\Rightarrow \dfrac{1+1}{2}=\dfrac{2}{2}=1$
Hence the answer to the above question is 1.

Note: We have to remember the cosh x formula to solve this question. In a hurry we can make a mistake by substituting $\dfrac{{{e}^{x}}-{{e}^{-x}}}{2}$ in place of $\dfrac{{{e}^{x}}+{{e}^{-x}}}{2}$ in equation (1) but we have to remember that $\dfrac{{{e}^{x}}-{{e}^{-x}}}{2}$ is formula of sinh x. Also we may substitute e to the power zero as 0 in a hurry so we need to be careful while doing these steps.