The value of \(\displaystyle \lim_{x \rightarrow \infty} (1+x^2)^{e^{-x}}\) is
2015
The value of \(\displaystyle \lim_{x \rightarrow \infty} (1+x^2)^{e^{-x}}\) is
- A.
\(0\) - B.
\(\frac {1} {2}\) - C.
\(1\) - D.
\(\infty\)
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Correct answer: C
Solution: We evaluate the limit by taking logarithms.
Write the expression as exp(e^{-x} * ln(1+x^2)). Then study the exponent e^{-x} * ln(1+x^2).
Consider the ratio ln(1+x^2)/e^{x}. Using L'Hôpital's rule, since both numerator and denominator tend to infinity, compute the derivative ratio: (2x/(1+x^2))/e^{x}.
The derivative ratio (2x/(1+x^2))/e^{x} tends to 0 because the exponential e^{x} grows much faster than the rational function 2x/(1+x^2). Thus ln(1+x^2)/e^{x} → 0.
Therefore e^{-x} * ln(1+x^2) = ln(1+x^2)/e^{x} → 0, so the original expression tends to exp(0) = 1.
Answer: 1