Use Taylor series to evaluate the following limits. limx→4ln⁡(x−3)x2−16

To evaluate the limit using Taylor series, limx→4ln⁡(x−3)x2−16 ∵ln⁡(1+x)=x−x22+x33−x44+... ∴ln⁡(x−3)=ln⁡(1+(x−4)) =(x−4)−(x−4)22+(x−4)33−(x−4)44+... limx→4ln⁡(x−4)x2−16=(x−4)−(x−4)22+(x−4)33−(x−4)44+...(x−4)(x+4) =(x−4)[1−(x−4)2+(x−4)23−(x−4)34+...](x−4)(x+4) 1−(x−4)2+(x−4)23−(x−4)34+...x+4 1−0+0−0+...4+4 =18 ∴limx→4ln⁡(x−3)x2−16=18

Explained: "Use Taylor series to evaluate the following limits...."

texelaare

Answered question

2020-11-22

Use Taylor series to evaluate the following limits.

lim_{x \to 4} \frac{\ln (x - 3)}{x^{2} - 16}

Answer & Explanation

Cullen

Skilled2020-11-23Added 89 answers

To evaluate the limit using Taylor series,

lim_{x \to 4} \frac{\ln (x - 3)}{x^{2} - 16}

∵ \ln (1 + x) = x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \frac{x^{4}}{4} + . . .

∴ \ln (x - 3) = \ln (1 + (x - 4))

= (x - 4) - \frac{(x - 4)^{2}}{2} + \frac{(x - 4)^{3}}{3} - \frac{(x - 4)^{4}}{4} + . . .

lim_{x \to 4} \frac{\ln (x - 4)}{x^{2} - 16} = \frac{(x - 4) - \frac{(x - 4)^{2}}{2} + \frac{(x - 4)^{3}}{3} - \frac{(x - 4)^{4}}{4} + . . .}{(x - 4) (x + 4)}

= \frac{(x - 4) [1 - \frac{(x - 4)}{2} + \frac{(x - 4)^{2}}{3} - \frac{(x - 4)^{3}}{4} + . . .]}{(x - 4) (x + 4)}

\frac{1 - \frac{(x - 4)}{2} + \frac{(x - 4)^{2}}{3} - \frac{(x - 4)^{3}}{4} + . . .}{x + 4}

\frac{1 - 0 + 0 - 0 + . . .}{4 + 4}

= \frac{1}{8}

∴ lim_{x \to 4} \frac{\ln (x - 3)}{x^{2} - 16} = \frac{1}{8}

Jeffrey Jordon

Expert2022-04-01Added 2605 answers

Answer is given below (on video)

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