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Ede4ka [16]
4 years ago
5

Lim x -> infinity ((e^(3x)) - (e^(-3x)))/( (e^(3x)) + (e^(-3x)))

Mathematics
1 answer:
steposvetlana [31]4 years ago
7 0
When x<span> approaches to </span><span>+∞</span><span> the function </span><span>e^<span>3x</span></span><span> becomes much bigger then </span><span>e^<span>−3x</span></span><span>, which obviously means that </span><span>e^<span>−3x</span></span><span> can be neglected in both numerator and denominator.
</span><span>
Here's how I figured this out: 

</span><span>lim <span>x →+∞ </span></span>= (<span><span><span>e^(<span>3x))</span></span>− (<span>e^(<span>−3x)) / (</span></span></span><span><span>e^<span>3x)) </span></span>+ (<span>e^(<span>−3x)) </span></span></span></span>= <span>lim <span>x → +∞ </span></span><span><span>e^<span>3x / </span></span><span>e^<span>3x </span></span></span>= 1
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