Explained: "Evaluate the given limits. a) limx→0(2sin⁡x+ex) b) limx→2x−2x2−4"

Tammy Todd

Answered question

2021-10-29

Evaluate the given limits.
a)

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

lim_{x \to 2} \frac{x - 2}{x^{2} - 4}

berggansS

Skilled2021-10-30Added 91 answers

Solution: Here we find the limits of follow:
a)

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

= lim_{x \to 0} 2 \sin x + lim_{x \to 0} e^{x}

= 2 lim_{x \to 0} \sin x + lim_{x \to 0} e^{x}

= 2 \cdot 0 + 1

= 1

Hence

lim_{x \to 0} (2 \sin x + e^{x}) = 1

lim_{x \to 2} \frac{x - 2}{x^{2} - 4}

= lim_{x \to 2} \frac{x - 2}{(x - 2) (x + 2)}

= lim_{x \to 2} \frac{1}{x + 2}

= \frac{1}{4}

Hence we get

lim_{x \to 2} \frac{x - 2}{x^{2} - 4} = \frac{1}{4}

Jeffrey Jordon

Expert2022-06-29Added 2605 answers

Answer is given below (on video)

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