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f(x)=(8x^2+x)/(2x^2+3)−6−5−4−3−2−10123456−3−2−101234567xy

Exercise

lim⁡x→∞(8x2+x2x2+3)\lim_{x\to\infty}\left(\frac{8x^2+x}{2x^2+3}\right)

Step-by-step Solution

1

As it's an indeterminate limit of type ∞∞\frac{\infty}{\infty}, divide both numerator and denominator by the term of the denominator that tends more quickly to infinity (the term that, evaluated at a large value, approaches infinity faster). In this case, that term is

lim⁡x→∞(8x2+xx22x2+3x2)\lim_{x\to\infty }\left(\frac{\frac{8x^2+x}{x^2}}{\frac{2x^2+3}{x^2}}\right)
2

Separate the terms of both fractions

lim⁡x→∞(8x2x2+xx22x2x2+3x2)\lim_{x\to\infty }\left(\frac{\frac{8x^2}{x^2}+\frac{x}{x^2}}{\frac{2x^2}{x^2}+\frac{3}{x^2}}\right)
3

Simplify the fraction

lim⁡x→∞(8+xx22+3x2)\lim_{x\to\infty }\left(\frac{8+\frac{x}{x^2}}{2+\frac{3}{x^2}}\right)
4

Simplify the fraction by xx

lim⁡x→∞(8+1x2+3x2)\lim_{x\to\infty }\left(\frac{8+\frac{1}{x}}{2+\frac{3}{x^2}}\right)
5

Evaluate the limit lim⁡x→∞(8+1x2+3x2)\lim_{x\to\infty }\left(\frac{8+\frac{1}{x}}{2+\frac{3}{x^2}}\right) by replacing all occurrences of xx by ∞\infty

44

Final answer to the exercise

44

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Main Topic: Limits by Direct Substitution

Find limits of functions at a specific point by directly plugging the value into the function.

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limx→∞(8x2+x2x2+3 )
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