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Compute the limit [tex]\lim_{x \to \ 0 } \frac{e^{2x}-1}{sin(x)}[/tex]

Sagot :

LammettHash

Both the numerator and denominator converge to e⁰ - 1 = 0 and sin(0) = 0. Applying L'Hopital's rule gives

[tex]\displaystyle \lim_{x\to0} \frac{e^{2x}-1}{\sin(x)} = \lim_{x\to0}\frac{2e^{2x}}{\cos(x)} = \frac{2e^0}{\cos(0)} = \boxed{2}[/tex]

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