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  • <img alt="\text{ Evaluate :}\mathrm{\lim _{x \rightarrow \infty} \frac{x+7 \sin x}{-2 x+13}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Ctext%7B%20Evaluate%20%3A%7D%5Cmathrm%7B%5Clim%2

\text{ Evaluate :}\mathrm{\lim _{x \rightarrow \infty} \frac{x+7 \sin x}{-2 x+13}}  using Sandwich Theorem.

Option: 1

\frac{1}{2}


Option: 2

\frac{-1}{2}


Option: 3

1


Option: 4

\frac{7}{11}


Answers (1)

best_answer

We have to evaluate \mathrm{\lim _{x \rightarrow \infty} \frac{x+7 \sin x}{-2 x+13}},

we know that \mathrm{\quad-1 \leq \sin x \leq 1} for all \mathrm{x}.

\mathrm{\Rightarrow -7 \leq 7 \sin x \leq 7 }
\mathrm{\Rightarrow x-7 \leq x+7 \sin x \leq x+7 }

Dividing throughout by \mathrm{(-2 x+13) }, we get
\mathrm{\frac{x-7}{(-2 x+13)} \geq \frac{x+7 \sin x}{(-2 x+13)} \geq \frac{x+7}{(-2 x+13)} \text { for all } x }.

Since, the quantity, \mathrm{\lim _{x \rightarrow \infty}(-2 x+13) } will be negative so inequality sign has changed.
\mathrm{\lim _{x \rightarrow \infty} \frac{x-7}{(-2 x+13)}=-\frac{1}{2} }
\mathrm{\lim _{x \rightarrow \infty} \frac{x+7 \sin x}{-2 x+13}=-\frac{1}{2} }

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Gunjita

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