Fwd: Ask My Instructor Question - Section 2.6, Question: 2.6.35

[William DeMeo]

---------- Forwarded message ----------
Date: Mon, Feb 2, 2015 at 11:48 PM
Subject: Ask My Instructor Question - Section 2.6, Question: 2.6.35

Book title: Thomas' Calculus Early Transcendentals, 12e
Course name: MATH_-165_-13
Objective: Find limits as x approaches infinity or negative infinity.

Optional -- Section 2.6 -- All MLP Exercises (Question # 18)

Question Link (this link will remain valid for 6 months from 02/02/15):
http://www.mathxl.com/info/exercise.aspx?fromask=yes&dataid=1733509e-953b-40ab-b7f1-fe1815f5aadd

Hello,

The doubt that I have is, can we bring the denominator from sq rt
16x^4+1 to 4x^2+1?
If not, please help me with this question.

Thanks,
Dhanush

[William DeMeo]

Dear Dhanush,

The answer to your question is no, \sqrt{16 x^4 + 1} is not equal to 4x^2 + 1.

(To check this compute (4x^2 + 1)^2 and see whether you get 16 x^4 + 1.) 

To solve this problem, first factor x^4 out of the expression inside the radical, as follows:

16 x^4 + 1 = x^4 (16 + 1/x^4)

Then the radical factors as

\sqrt{ x^4 (16 + 1/x^4) } = \sqrt{ x^4 } \sqrt{16 + 1/x^4} = x^2 \sqrt{16 + 1/x^4}

This is good news because there's an x^2 in the numerator, so factor x^2 out of both the numerator and denominator and cancel them to arrive at

       3 - 1/x^2
-----------------------
\sqrt{ 16 + 1/x^4 }

The limit of this expression, as x tends to infinity, is 3/\sqrt{16} or 3/4.

Sincerely,
William