Question

Problem 3: Find the following limits (may or may not exist) (a) (6 points) $\lim_{x \to 0} \frac{e^x - 1 - x - \frac{x^2}{2}}{x^3}$ (b) (6 points) $\lim_{x \to 0^+} x^{2x}$ (c) (6 points) $\lim_{x \to 0} \frac{e^{-\frac{1}{x^2}}}{x^2}$ (d) (6 points) $\lim_{x \to 0^+} x \sin(\frac{1}{x})$ (e) (6 points) $\lim_{x \to 2^+} \frac{\sqrt{x^2 - 4}}{\sqrt{x^3 - 8}}$ (d) (6 points) Determine if the series converges: $\sum_{k=2}^{k=\infty} \frac{1}{(\ln k)k^{1.5}}$

          Problem 3: Find the following limits (may or may not exist)
(a) (6 points) $\lim_{x \to 0} \frac{e^x - 1 - x - \frac{x^2}{2}}{x^3}$
(b) (6 points) $\lim_{x \to 0^+} x^{2x}$
(c) (6 points) $\lim_{x \to 0} \frac{e^{-\frac{1}{x^2}}}{x^2}$
(d) (6 points) $\lim_{x \to 0^+} x \sin(\frac{1}{x})$
(e) (6 points) $\lim_{x \to 2^+} \frac{\sqrt{x^2 - 4}}{\sqrt{x^3 - 8}}$
(d) (6 points) Determine if the series converges: $\sum_{k=2}^{k=\infty} \frac{1}{(\ln k)k^{1.5}}$

Problem 3: Find the following limits (may or may not exist)
(a) (6 points) limx → 0(e^x - 1 - x - (x^2)/(2))/(x^3)
(b) (6 points) limx → 0^+ x^2x
(c) (6 points) limx → 0(e^-(1)/(x^2))/(x^2)
(d) (6 points) limx → 0^+ x sin((1)/(x))
(e) (6 points) limx → 2^+(√(x^2 - 4))/(√(x^3 - 8))
(d) (6 points) Determine if the series converges: ∑k=2^k=∞(1)/((ln k)k^1.5)

Added by Aaron R.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Vincenzo Zaccaro

08/19/2023

Step 1

We can simplify the expression by using Taylor series expansion: e^x = 1 + x + (x^2)/2 + O(x^3) Substitute this into the expression: (e^(x)-1-x-(x^(2))/(2))/(x^(3)) = ((1 + x + (x^2)/2 + O(x^3)) - 1 - x - (x^2)/2)/(x^3) = (x^2/2 + O(x^3))/(x^3) = 1/2 + O(x) As Show more…

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Problem 3: Find the following limits (may or may not exist) (a) (6 points) lim_(x->0)(e^(x)-1-x-(x^(2))/(2))/(x^(3)). (b) (6 points) lim_(x->0^(+))x^(2x^(x)) (c) (6 points) lim_(x->0)((-1)/(x^(2)))/(x^(2)) (d) (6 points) lim_(x->0^(+))xsin((1)/(x)) (e) (6 points) lim_(x->2^(+))(sqrt(x^(2)-4))/(sqrt(x^(3)-8)) (d) (6 points) Determine if the series converges: sum_(k=2)^(k=infty ) (1)/((lnk)k^(1.5)). x2 23 (a) (6 points) limx->0 ln (c) (6 points) limx->0 2 (d) (6 points) limx->0 +xsin() (e) (6 points) limx->2+ OX (lnk)k^1.5 Quk