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Question: Let f(x) = g(x) \(\frac{e^{1/x} - e^{- 1/x}}{e^{1/x} + e^{- 1/x}}\)and x ¹ 0 where g is a continuous...

Let f(x) = g(x) e1/x−e−1/xe1/x+e−1/x\frac{e^{1/x} - e^{- 1/x}}{e^{1/x} + e^{- 1/x}}and x ¹ 0 where g is a continuous function. Then lim⁡x→0\lim_{x \rightarrow 0}f(x) exists if

A

g(x) is any polynomial

B

g(x) = x + 4

C

g(x) = x2

D

g(x) = 2 + 3x + 4x2

Answer

g(x) = x2

Explanation

Solution

lim⁡x→0+e1/x−e−1/xe1/x+e−1/x\lim_{x \rightarrow 0^{+}}\frac{e^{1/x} - e^{- 1/x}}{e^{1/x} + e^{- 1/x}} = lim⁡x→0+1−e−2/x1+e−2/x\lim_{x \rightarrow 0^{+}}\frac{1 - e^{- 2/x}}{1 + e^{- 2/x}}= 1 and

lim⁡x→0−e1/x−e−1/xe1/x+e−1/x\lim_{x \rightarrow 0^{-}}\frac{e^{1/x} - e^{- 1/x}}{e^{1/x} + e^{- 1/x}}= lim⁡x→0−e2/x−1e2/x+1\lim_{x \rightarrow 0^{-}}\frac{e^{2/x} - 1}{e^{2/x} + 1}= –1. Hence

lim⁡x→0f(x)\lim_{x \rightarrow 0}f(x) exists if g(x) = x or g(x) = x2. If g(x) = a (a ¹ 0), thenlim⁡x→0+f(x)\lim_{x \rightarrow 0 +}f(x) = a and lim⁡x→0−f(x)\lim_{x \rightarrow 0 -}f(x) = –a. Thus lim⁡x→0f(x)\lim_{x \rightarrow 0}f(x) doesn't exist in this case.