Calculate $\lim_{x \to 0} (\cos x)^{1/x^2}$.
Let L = $\lim (cos x)^{1/x^2}$. ln L = $\lim \frac{\ln(\cos x)}{x^2} = \lim \frac{-\tan x}{2x} = -1/2$. Thus $L = e^{-1/2$} $\approx 0.6065 \approx 0.61$.
If $f(x) = x^2$ $\sin(1/x)$ for x
eq 0 and f(0) = 0, which of the following statements is TRUE?
f'(0) = $\lim_{h \to 0} (h^2 \sin(1/h) - 0)/h = \lim_{h \to 0} h \sin(1/h) = 0$. So f is differentiable at x=0, though f'(x) fluctuates wildly near 0.
Evaluate $\lim_{x \to 0} \frac{x \tan x}{1 - \cos x}$.
Using Taylor expansions: x tan x $\approx x^2, 1 - cos x \approx x^2/2$. Limit = $x^2 / (x^2/2)$ = 2.
Which of the following functions are continuous everywhere on $\mathbb{R}$?
The floor function \lfloor x
floor has jump discontinuities at all integers. The other three functions are continuous everywhere.
If f(x) = $\begin{cases} a x + b & x \le 1 \\ x^2 + 3x + 2 & x > 1 \end{cases}$ is differentiable at x = 1, find the value of a + b.
For continuity at x=1: a + b = 1 + 3 + 2 = 6. For differentiability at x=1: a = 2(1) + 3 = 5. Hence b = 1, and a + b = 6.
