Explained: "Solving the following limit without LHospitals rule: limx→0sin⁡(x2+2)−sin⁡(x+2)x" - Plainmath

gea3stwg

Answered question

2022-01-24

Solving the following limit without LHospitals rule:

lim_{x \to 0} \frac{\sin (x^{2} + 2) - \sin (x + 2)}{x}

Answer & Explanation

caoireoilns

Beginner2022-01-25Added 12 answers

You can just sum-to-product it from the beginning and use

lim_{x \to 0} \frac{\sin x}{x} = 1

lim_{x \to 0} \frac{\sin (x^{2} + 2) - \sin (x + 2)}{x} = lim_{x \to 0} \frac{2 \cos [2 + x \frac{x + 1}{2}] \sin [x \frac{x - 1}{2}]}{x}

= \cos (2) lim_{x \to 0} \frac{\sin [x \frac{x - 1}{2}]}{\frac{x}{2}}

= - \cos (2) lim_{x \to 0} \frac{\sin [x \frac{x - 1}{2}]}{x \frac{x - 1}{2}}

= - \cos (2)

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