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  • Find     <img alt="\mathrm{\lim _{x \rightarrow \infty} \frac{\sin x+\cos x}{x}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Clim%20_%7Bx%20%5Crightarrow%20%5

Find     \mathrm{\lim _{x \rightarrow \infty} \frac{\sin x+\cos x}{x}}

Option: 1

0


Option: 2

1


Option: 3

2


Option: 4

-2


Answers (1)

best_answer

Yes thats simple the squeze Theorem, which states if  \mathrm{g(x)<f(x)<h(x)} as X approaches \mathrm{a} and

\mathrm{g(x)=h(x)=L}, then:

                                             \mathrm{\begin{aligned} & \lim _{x \rightarrow a} f(x)=L \\\\ & -1 \leq \sin x \leq 1 \\\\ & -1 \leq \cos x \leq 1 \end{aligned}}

                        \mathrm{\begin{gathered} \Longrightarrow-2 \leq \sin x+\cos x \leq 2 \\\\ \Longrightarrow-\frac{2}{x} \leq \frac{\sin x+\cos x}{x} \leq \frac{2}{x} \\\\ \Longrightarrow \lim _{x \rightarrow \infty}-\frac{2}{x} \leq \lim _{x \rightarrow \infty} \frac{\sin x+\cos x}{x} \leq \lim _{x \rightarrow \infty} \frac{2}{x} \\\\ \Longrightarrow 0 \leq \lim _{x \rightarrow \infty} \frac{\sin x+\cos x}{x} \leq 0 \end{gathered}}

Therefore the limit of the function is 0.

                                     \mathrm{\lim _{x \rightarrow \infty}\left(\frac{\sin x+\cos x}{x}\right)=0}

 

Posted by

Nehul

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