Question

i) $\lim_{x \to \infty} \frac{x + x^2}{2 - 3x^2}$ ii) If $f'$ is continuous and $f(2) = f'(2) = 0$, and $f''(2) = 4$, compute $\lim_{x \to 0} \frac{f(2 + 3x) + f(2 + 5x)}{x^2}$ iii) $\lim_{x \to \infty} x(\frac{\pi}{2} - \arctan x)$ iv) $\lim_{x \to 0^+} (\frac{1}{x} - \frac{1}{e^x - 1})$ v) $\lim_{x \to \infty} x e^{-x}$

          i) $\lim_{x \to \infty} \frac{x + x^2}{2 - 3x^2}$

ii) If $f'$ is continuous and $f(2) = f'(2) = 0$, and $f''(2) = 4$, compute
$\lim_{x \to 0} \frac{f(2 + 3x) + f(2 + 5x)}{x^2}$
iii) $\lim_{x \to \infty} x(\frac{\pi}{2} - \arctan x)$
iv) $\lim_{x \to 0^+} (\frac{1}{x} - \frac{1}{e^x - 1})$
v) $\lim_{x \to \infty} x e^{-x}$

i) limx →∞(x + x^2)/(2 - 3x^2)

ii) If f' is continuous and f(2) = f'(2) = 0, and f”(2) = 4, compute
limx → 0(f(2 + 3x) + f(2 + 5x))/(x^2)
iii) limx →∞ x((π)/(2) - arctan x)
iv) limx → 0^+ ((1)/(x) - (1)/(e^x - 1))
v) limx →∞ x e^-x

Added by Kathy B.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Satyam Gupta

11/04/2021

Step 1

To simplify the expression, we substitute the given function $f(x) = x + x^2 - 2 - 3x^2$ into the expression. $f(x+3x) + f(x+5x) = (x+3x) + (x+3x)^2 - 2 - 3(x+3x)^2 + (x+5x) + (x+5x)^2 - 2 - 3(x+5x)^2$ Simplifying further, we get: \(f(x+3x) + f(x+5x) = 4x Show more…

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Texts: If (f(x) = x + x^2 - 2 - 3x^2) is continuous and (f(2) = 0) and (f'(2) = 4), compute (lim_{x o 0} frac{f(x+3x) + f(x+5x)}{x^2}). ii) (lim_{x o -arctan(T)} 2). iv) (lim_{x o +0} x). v) (lim_{x o e}).