$\lim_{{x} \to {-\infty}} \left(\frac{x+a}{x+b}\right)^{a+b}$ is equal to

$\lim_{{x} \to {-\infty}} \left(\frac{x+a}{x+b}\right)^{a+b}$ is equal to
  1. $1$
  2. $e^{b-a}$
  3. $e^{a-b}$
  4. $e^b$

Solution

$\lim _{x \rightarrow \infty}\left(\frac{x+a}{x+b}\right)^{a+b}=\lim _{x \rightarrow \infty}\left(\frac{1+\frac{a}{x}}{1+\frac{b}{x}}\right)^{a+b}=(1)^{a+b}=1$

Asked in: AP EAMCET 2001

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