Answer to Question #91363 in Calculus for Sajid

Question

Q. Choose the correct answer.
Q. lim┬(n→∞)⁡□((a^n-a^(-n))/(a^n+a^(-n) )) =?
a. 0
b. 1
c. -1
d. ∞

Answer 1

$$\lim_{n \to \infty }\frac{a^n-a^{-n}} {a^n+a^{-n}}=?$$

1) If a> 1

$$\lim_{n \to \infty }\frac{a^n-a^{-n}} {a^n+a^{-n}}=\lim_{n \to \infty }\frac{a^n(1-a^{-2n})} {a^n(1+a^{-2n})}=\lim_{n \to \infty }\frac{(1-a^{-2n})} {(1+a^{-2n})}=1$$

The correct answer :b 1 

2) If 0<a< 1

$$\lim_{n \to \infty }\frac{a^n-a^{-n}} {a^n+a^{-n}}=\lim_{n \to \infty }\frac{(1/a)^n((1/a)^{-2n}-1)} {(1/a)^n((1/a)^{-2n}+1 )}=\lim_{n \to \infty }\frac{((1/a)^{-2n}-1)} {((1/a)^{-2n}+1)}=-1$$

the correct answer: c -1  .

3) If a=1

$$\lim_{n \to \infty }\frac{a^n-a^{-n}} {a^n+a^{-n}}=\lim_{n \to \infty }\frac{1^n- 1^{-n}} {1^n+1^{-n}}=\lim_{n \to \infty }\frac{0} {2}=0$$

the correct answer: a 0

4) If a= - 1

$$\lim_{n \to \infty }\frac{a^n-a^{-n}} {a^n+a^{-n}}=\lim_{n \to \infty }\frac{(-1)^n-(-1) ^{-n}} {(-1)^n+(-1)^{-n}}=\lim_{n \to \infty }\frac{(-1)^{2n}-1} {(-1)^{2n}+1}=\lim_{n \to \infty }\frac{1-1} {1+1}=0$$

5) If -1<a< 0

$$\lim_{n \to \infty }\frac{a^n-a^{-n}} {a^n+a^{-n}}=\lim_{n \to \infty }\frac{((-1(-a))^n-(-1(-a))^{-n}} {(-1(-a))^n+(-1(-a))^{-n}}=\lim_{n \to \infty }\frac{(1/a)^n((1/a)^{-2n}-1)} {(1/a)^n((1/a)^{-2n}+1 )}$$

We will introduce a replacement b=-a (0<b< 1 )

$$\lim_{n \to \infty }\frac{((-1(b))^n-(-1(b)^{-n}} {(-1(b))^n+(-1(b))^{-n}}=\lim_{n \to \infty }\frac{(-1)^n(b^n-b^{-n})} {(-1)^n(b^n+b^{-n})}=\\ \\ =\lim_{n \to \infty }\frac{(b^n-b^{-n})} {(b^n+b^{-n})}=\lim_{n \to \infty }\frac{(1/b)^n((1/b)^{-2n}-1)} {(1/b)^n((1/b)^{-2n}+1 )} = -1$$

the correct answer: c -1.

6) If a< - 1

If we make b= -a ,( b> 1 ) then we get a sequence

$$\lim_{n \to \infty }\frac{a^n-a^{-n}} {a^n+a^{-n}}=\lim_{n \to \infty }\frac{(-1)^n b^n(1-b^{-2n})} {(-1)^n b^n(1+b^{-2n})}=\lim_{n \to \infty }\frac{(1-b^{-2n})} {(1+b^{-2n})}=1$$

the correct answer :. b 1 

7) If a=0

The sequence $z_n=\frac{a^n-a^{-n}} {a^n+a^{-n}}$ is undefined (division by zero).