Answered

Looking for reliable answers? Westonci.ca is the ultimate Q&A platform where experts share their knowledge on various topics. Get immediate and reliable answers to your questions from a community of experienced experts on our platform. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.

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]

We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.