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Compute the limit [tex]\lim_{n \to \ 0} \frac{log(1+x)}{cos(x)+e^{x} -1}[/tex]

Sagot :

LammettHash

I assume you mean

[tex]\displaystyle \lim_{x\to0} \frac{\log(1+x)}{\cos(x) + e^x - 1}[/tex]

since the limand is free of n. As x goes to 0, the numerator converges to log(1 + 0) = 0, while the denominator converges to cos(0) + e⁰ - 1 = 1, so the overall limit is

[tex]\displaystyle \lim_{x\to0} \frac{\log(1+x)}{\cos(x) + e^x - 1} = \frac01 = \boxed{0}[/tex]

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