Can someone explain how to do this question [Limits Question] (1 Viewer)

Jig · Jan 23, 2019

The answer is y=1 but i don't understand the worked solutions to get there.

integral95 · Jan 23, 2019

In order to find the horizontal asymptote, you need to take the limit as the x-value approaches infinity. You also need use this concept.

$lim_{x \rightarrow 0} \frac{sinx}{x} =lim_{x \rightarrow \infty} \frac{sin(\frac{1}{x})}{\frac{1}{x}} = 1$

$lim_{x \rightarrow \infty} \frac {x^2 sin(\frac{1}{x})}{x+1} \\ \\ = lim_{x \rightarrow \infty}\frac{sin(\frac{1}{x})}{\frac{1}{x}} \frac{x}{x+1} \\ \\ \Rightarrow 1*1 = 1$

Jig · Jan 24, 2019

integral95 said:
In order to find the horizontal asymptote, you need to take the limit as the x-value approaches infinity. You also need use this concept.

$lim_{x \rightarrow 0} \frac{sinx}{x} =lim_{x \rightarrow \infty} \frac{sin(\frac{1}{x})}{\frac{1}{x}} = 1$

$lim_{x \rightarrow \infty} \frac {x^2 sin(\frac{1}{x})}{x+1} \\ \\ = lim_{x \rightarrow \infty}\frac{sin(\frac{1}{x})}{\frac{1}{x}} \frac{x}{x+1} \\ \\ \Rightarrow 1*1 = 1$

Ahh Okay, thanks.

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Can someone explain how to do this question [Limits Question] (1 Viewer)

Jig

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integral95

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Jig

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