"Find the limits. a) limx→∞x2+5x−x b) limx→∞sin⁡(πx2−3x)" was answered by students like you

geduiwelh

Answered question

2021-10-17

Find the limits.
a)

lim_{x \to \infty} \sqrt{x^{2} + 5 x} - x

lim_{x \to \infty} \sin (\frac{π x}{2 - 3 x})

unett

Skilled2021-10-18Added 119 answers

Given:
a)

lim_{x \to \infty} \sqrt{x^{2} + 5 x} - x

lim_{x \to \infty} \sin (\frac{π x}{2 - 3 x})

Calculation:
Multiple and divide the conjugate of

\sqrt{x^{2} + 5 x} - x

\frac{\sqrt{x^{2} + 5 x} - x}{\sqrt{x^{2} + 5 x} + x} \sqrt{x^{2} + 5 x} + x = \frac{\sqrt{{(x^{2} + 5 x)}^{2}} - x^{2}}{\sqrt{x^{2} + 5 x} + x}

= \frac{5 x}{\sqrt{x^{2} + 5 x} + x}

Divided by highest denominator power,

= 5 lim_{x \to \infty} \frac{x}{\sqrt{1 + \frac{5}{x}} + 1}

= 5 \frac{1}{2}

lim_{x \to \infty} \sin (\frac{π x}{2 - 3 x}) = lim_{x \to \infty} \sin (\frac{π x}{x (\frac{2}{x} - 3)})

lim_{x \to \infty} \sin (\frac{π}{\frac{2}{x} - 3})

= \sin (- \frac{π}{3})

= - \frac{\sqrt{3}}{2}

Answer:
a)

lim_{x \to \infty} \sqrt{x^{2} + 5 x} - x = \frac{5}{2}

lim_{x \to \infty} \sin (\frac{π x}{2 - 3 x}) = \frac{3}{2}

Jeffrey Jordon

Expert2022-06-29Added 2605 answers

Answer is given below (on video)

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