Let . Which of the following best explains why approaches different finite values as…

MATHEMATICS · Single correct · A Level · Mathematics · Exponentials and logarithms · Exponentials and logarithms

Problem

Let

g(x)=\dfrac{e^{x}-e^{-x}}{e^{x}+e^{-x}}

. Which of the following best explains why

g

approaches different finite values as

x\to\infty

and

x\to-\infty

? (a) Dividing numerator and denominator by

e^{x}

shows

g(x)\to 1

x\to\infty

, and the same division works for

x\to-\infty

, so the ratio approaches the single constant

1

on both ends. (b) For large positive

x

e^{-x}

becomes negligible leaving

\displaystyle g(x)=\frac{-e^{-x}}{e^{-x}}\approx-1

, while for large negative

x

the roles switch giving a limit of

1

, so the limits are

-1

and

1

respectively. (c) For large positive

x

e^{x}

dominates both numerator and denominator so the ratio approaches

1

, whereas for large negative

x

e^{-x}

dominates, making the ratio approach

-1

. (d) Because

e^{x}

and

e^{-x}

grow without bound in opposite directions, their difference is small compared with their sum, causing the ratio to approach

0

at both extremes.