Let's do some calculus! (60)

Calculus Level 3

$\large \lim_{x \to - \infty} \left(\sqrt[5]{x^5 + 7x^4 + 2} - x\right) =\ ?$

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The answer is 1.4.

2 solutions

Chew-Seong Cheong
Jun 5, 2018

$\begin{aligned} L & = \lim_{x \to -\infty} \sqrt[5]{x^5+7x^4+2} - x \\ & = \lim_{x \to -\infty} x {\color{#3D99F6}\left(1+\frac 7x+\frac 2{x^5}\right)^\frac 15} - x & \small \color{#3D99F6} \text{By Taylor series expansion} \\ & = \lim_{x \to -\infty} x\left(1+ \frac 15\left(\frac 7x+\frac 2{x^5}\right) - \frac 2{25} \left(\frac 7x+\frac 2{x^5}\right)^2 + \cdots \right) - x \\ & = \lim_{x \to -\infty} \left(x+ \frac 75 + \frac {98}{25x} + \cdots \right) - x \\ & = \frac 75 = \boxed{1.4} \end{aligned}$

Good solution! Btw is it Taylor expansion or simply Binomial expansion?

Tapas Mazumdar - 3 years ago

I think either name is okay.

Chew-Seong Cheong - 3 years ago

@Tapas Mazumdar I had the same doubt in some other problem...............actually, if you derive the formula of the Binomial Expansion, using the method used in finding the Taylor series, you will get the same thing..........And we can see ( with some effort ) that the Remainder tends to 0 and hence, the Binomial expansion is equal to its Taylor series expansion.....!!

Aaghaz Mahajan - 2 years, 12 months ago

Aaghaz Mahajan
Jun 1, 2018

Replace x by 1/y.......And then, use fractional Binomial theorem....!!!

Let's do some calculus! (60)

For more problems on calculus, click here .

The answer is 1.4.

2 solutions

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