Always be in your limits

Calculus Level 3

$\large \lim _{ x\to\infty } \left[ \sqrt[3]{ (x+3)(x+5)(x+7) } -x \right] = \ ?$

The answer is 5.

1 solution

Luis Valdivia
Oct 3, 2015

Notice that when multiplied, the terms under the root begin with $x^3+15x^2+...$ , which resembles $(x+5)^3$ when distributed. For such a case, as x approaches infinity, we only care about the first two terms of the expression under the root. Thus, we can say that the expression under the root is approximately equal to $(x+5)^3.$ Substituting that under the root and implifying, we get $x+5-x$ , which can be further reduced to 5. Therefore, as $x \rightarrow \infty$ , 5 = 5.

True in this case but what if there is

(x+a)(x+b)(x+c)

Prakher Gaushal - 5 years, 8 months ago

Then the expression under the expression can be approximated to $(x+(\cfrac{a+b+c}{3})^3$

The expression under the root generally can be approximated to $(x+m)^n$ where m the average of the constant parts(in this case a, b, and c) and n is the number of first degree binomials present.

Luis Valdivia - 5 years, 8 months ago

Always be in your limits

The answer is 5.

1 solution

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