complex number question

\(\mathop {\lim }\limits_{x \to \infty } \dfrac{{6{e^x} - 15{e^{ - x}}}}{{2{e^x} - 3{e^{ - x}}}} = \mathop {\lim }\limits_{x \to \infty } \dfrac{{6{e^{2x}} - 15}}{{2{e^{2x}} - 3}}\overbrace = ^H\mathop {\lim }\limits_{x \to \infty } \dfrac{{12{e^{2x}}}}{{4{e^{2x}}}} = 3\)
 
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