What is the limit of tanh x as x approaches infinity?

[merced]

\lim_{x\rightarrow\infty} tanh x = \lim_{x\rightarrow\infty} \frac{sinh x}{cosh x}
= \lim_{x\rightarrow\infty} \frac{e^x - e^{-x}}{e^x + e^{-x}}
= \lim_{x\rightarrow\infty} \frac{e^{2x} -1}{e^{2x} +1}

what now?

 

[AKG]

Evaluate the limit.

 

[benorin]

\lim_{x\rightarrow\infty} tanh x = \lim_{x\rightarrow\infty} \frac{sinh x}{cosh x}
= \lim_{x\rightarrow\infty} \frac{e^x - e^{-x}}{e^x + e^{-x}}
= \lim_{x\rightarrow\infty} \frac{e^{2x} -1}{e^{2x} +1}

what now?

=^{H} \lim_{x\rightarrow\infty} \frac{2e^{2x}}{2e^{2x}}=1

where the symbol =^{H} denotes the use of l'Hospital's rule.

 

[merced]

thank you.

 

[HallsofIvy]

Swatting a fly with a sledgehammer!
Multiply both numerator and denominator of
= \lim_{x\rightarrow\infty} \frac{e^x - e^{-x}}{e^x + e^{-x}}
by e^-x rather than e^x and you get
= \lim_{x\rightarrow\infty} \frac{1- e^{-2x}}{1+ e^{-2x}}
which is obvious.