Question

Lim x tends to 1 (2x^8-3x^2 1)/(x^8 6x^5-7)

Answer 1

answer is 5/19

we have to evaluate Lim (x→ 1) (2x^8 - 3x² + 1)/(x^8 + 6x^5 - 7)

putting, x = 1 we get, (2 × 1 - 3 × 1 + 1)/(1 + 6 - 7) = 0/0 . so it is in the form of limit.

now we can apply L- Hospital rule,

differentiating numerator and denominator separately,

lim(x→1) (2 × 8x^7 - 6x)/(8x^7 + 6 × 5x⁴ )

= lim(x → 1) (16x^7 - 6x)/(8x^7 + 30x⁴)

now putting x = 1 we get,

= (16 × 1^7 - 6 × 1)/(8 × 1^7 + 30 × 1⁴)

= (16 - 6)/(8 + 30)

= 10/38

= 5/19

hence value of given limit is 5/19

also read similar questions : lim x→ 3 (2x^2+3x+1) / (x+2)

https://brainly.in/question/13294274

lim x tends to 0 (1-cos^3x)/xsin2x

https://brainly.in/question/1150387