Question

(a) lim x --> 2 (2x^2-8)/(5x^2-7x-6) (b) lim x --> 2 (6x^4-3x^3-18x^2)/(x^4-2x^2-8) (c) lim x --> infinity (x^2+sqrt(5x^4+x^2+1))/(3x^2-x-sqrt(x^6+4x+9))

          (a) lim x --> 2   (2x^2-8)/(5x^2-7x-6)
(b) lim x --> 2   (6x^4-3x^3-18x^2)/(x^4-2x^2-8)
(c) lim x --> infinity   (x^2+sqrt(5x^4+x^2+1))/(3x^2-x-sqrt(x^6+4x+9))

Added by Jennifer L.

Calculus: Early Transcendentals

James Stewart 8th Edition

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Solved by Expert Rakhshan S

03/20/2022

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### Part (a) lim x --> 2 (2x^2-8)/(5x^2-7x-6) ** Show more…

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(a) lim x --> 2 (2x^2-8)/(5x^2-7x-6) (b) lim x --> 2 (6x^4-3x^3-18x^2)/(x^4-2x^2-8) (c) lim x --> infinity (x^2+sqrt(5x^4+x^2+1))/(3x^2-x-sqrt(x^6+4x+9))

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Transcript

00:01 We have been given the following three limits and we are supposed to solve those limits.

00:06 Evaluate the value of the limits.

00:08 Okay, so i'm solving the first part.

00:10 So you can see clearly if i put x equal to 2, we will get this in 0 by 0 form.

00:16 Okay.

00:17 So the first limit on putting x equal to 2 we are getting this as to be 0 by 0 form.

00:22 So i can apply l hopital rule.

00:24 So i am applying here l hopital rule.

00:31 So i will get my limit to be equal to limit x tends to 2.

00:37 So now take the derivative, so it becomes 4x minus 0 divided by 10x minus 7.

00:44 Okay, so this will be nothing but.

00:46 Now put x equal to 2, you will get 4 times 2 divided by 10 times 2 minus 7.

00:53 So this is nothing but 8 divided by 20 minus 7 is just 13.

00:58 So the first part answer will be 8 divided by 13.

01:04 Now let us see the second part.

01:06 Okay.

01:08 So in the second part also if you put x equal to 2, you will get 0 by 0 form.

01:13 Again we can apply the l -hopital rule.

01:16 So for the second part also, i am applying this l -hopital rule...

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