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Hint: We can easily solve problems like these by properly using the principles of limits. In the given problem we are required to find the limit of cosine of the expression $\dfrac{1}{x}$ when the variable $x$ tends to infinity. As the function cosine is a continuous one we can take the limit inside the function. After taking the limit inside the trigonometric function we see that the fraction $\dfrac{1}{x}$ approaches the value $0$ . From this, we deduct that the value of $\cos \left( 0 \right)$ is zero.
Complete step by step solution:
In the problem we are required to find the value of the limit $\cos \left( \dfrac{1}{x} \right)$ as the variable $x$ approaches infinity i.e., we need to find $\displaystyle \lim_{x \to \infty }\cos \left( \dfrac{1}{x} \right)=?$
In mathematics, the cosine function (generally known as $\cos $ function) in a triangle represents the ratio of the adjacent side to that of the hypotenuse. The cosine function is one of the three main primary trigonometric functions. The graph of the cosine function is always continuous.
According to the principles of limits we can move the limit inside the trigonometric function as we already know that the trigonometric cosine function is a continuous one.
Hence, $\displaystyle \lim_{x \to \infty }\cos \left( \dfrac{1}{x} \right)=\cos \left( \displaystyle \lim_{x \to \infty }\dfrac{1}{x} \right)$
Since the fraction $\dfrac{1}{x}$ has a real number in the numerator while its denominator is unbounded, the value of the fraction $\dfrac{1}{x}$ approaches $0$ .
Hence, $\cos \left( \displaystyle \lim_{x \to \infty }\dfrac{1}{x} \right)=\cos \left( 0 \right)$
We know that the value of $\cos \left( 0 \right)$ is $1$.
Therefore, the value of the required given is $1$.
Note: To solve this type of problem we must properly know the nature of the trigonometric functions, as we might not be able to incorporate the properties of limits properly without knowing the nature of the functions. Also, we have to be very careful while working with the value inside the cosine function, as it would be a totally different scenario if it was $\cos \left( x \right)$ instead of $\cos \left( \dfrac{1}{x} \right)$ .
Complete step by step solution:
In the problem we are required to find the value of the limit $\cos \left( \dfrac{1}{x} \right)$ as the variable $x$ approaches infinity i.e., we need to find $\displaystyle \lim_{x \to \infty }\cos \left( \dfrac{1}{x} \right)=?$
In mathematics, the cosine function (generally known as $\cos $ function) in a triangle represents the ratio of the adjacent side to that of the hypotenuse. The cosine function is one of the three main primary trigonometric functions. The graph of the cosine function is always continuous.
According to the principles of limits we can move the limit inside the trigonometric function as we already know that the trigonometric cosine function is a continuous one.
Hence, $\displaystyle \lim_{x \to \infty }\cos \left( \dfrac{1}{x} \right)=\cos \left( \displaystyle \lim_{x \to \infty }\dfrac{1}{x} \right)$
Since the fraction $\dfrac{1}{x}$ has a real number in the numerator while its denominator is unbounded, the value of the fraction $\dfrac{1}{x}$ approaches $0$ .
Hence, $\cos \left( \displaystyle \lim_{x \to \infty }\dfrac{1}{x} \right)=\cos \left( 0 \right)$
We know that the value of $\cos \left( 0 \right)$ is $1$.
Therefore, the value of the required given is $1$.
Note: To solve this type of problem we must properly know the nature of the trigonometric functions, as we might not be able to incorporate the properties of limits properly without knowing the nature of the functions. Also, we have to be very careful while working with the value inside the cosine function, as it would be a totally different scenario if it was $\cos \left( x \right)$ instead of $\cos \left( \dfrac{1}{x} \right)$ .
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