Question:

$\displaystyle \lim_{x \to 0}$ $\left(cos\,x+sin\,x\right)^{\frac{1}{x}}$ equals

Updated On: Jul 6, 2022
  • $e^1$
  • $e^{-1}$
  • $1$
  • $0$
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The Correct Option is A

Solution and Explanation

$\displaystyle \lim_{x \to 0}$ $\left(cos\,x+sin\,x\right)^{\frac{1}{x}}\quad(1^{\infty}$ form$)$ $=e^{\displaystyle \lim_{x \to 0}\frac{\left(cos\,x+sin\,x-1\right)}{x}}$ $=e^{\displaystyle \lim_{x \to 0}\left[\frac{sin\,x}{x}-\left(\frac{1-cos\,x}{x}\right)\right]}$ $=e^{1-0}=e^{1}$
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Concepts Used:

Limits And Derivatives

Mathematically, a limit is explained as a value that a function approaches as the input, and it produces some value. Limits are essential in calculus and mathematical analysis and are used to define derivatives, integrals, and continuity.

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derivative is referred to the instantaneous rate of change of a quantity with response to the other. It helps to look into the moment-by-moment nature of an amount. The derivative of a function is shown in the below-given formula.

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