First slide

Introduction to limits

Question

$L = lim_{x \to 0} \frac{\sin x + {ae}^{x} + {be}^{- x} + cln (1 + x)}{x^{3}} = \infty$

Moderate

Question

The value of L is

A

1 /2
B

-1 /3
C

-1/6
D

3

Solution

$L = lim_{x \to 0} \frac{\sin x + {ae}^{x} + {be}^{- x} + cln (1 + x)}{x^{3}}$
$= lim_{x \to 0} \frac{(x - \frac{x^{3}}{3!}) + a (1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \frac{x^{3}}{3!}) + b (1 - \frac{x}{1!} + \frac{x^{2}}{2!} - \frac{x^{3}}{3!}) + c (x - \frac{x^{2}}{2} + \frac{x^{3}}{3})}{x^{3}}$
$= lim_{x \to 0} \frac{(a + b) + (1 + a - b + c) x + (\frac{a}{2} + \frac{b}{2} - \frac{c}{2}) x^{2} + (- \frac{1}{3!} + \frac{a}{3!} - \frac{b}{3!} + \frac{c}{3}) x^{3}}{x^{3}}$
or $a + b = 0, 1 + a - b + c = 0, \frac{a}{2} + \frac{b}{2} - \frac{c}{2} = 0$
and $L = - \frac{1}{3!} + \frac{a}{3!} - \frac{b}{3!} + \frac{c}{3}$
Solving the first three equations, we get c = 0, a = -1/2, b = 1/2.
Then, L = - 1/3.
Equation ax² + bx + c = 0 reduces to x² -x=0 or x=0, 1.
$\begin{array}{l} | | x + c | - 2 a | < 4 b reduces to \\ ∥ x | + 1 | < 2 \\ or - 2 < | x | + 1 < 2 \\ or 0 \leq | x | < 1 \\ or x \in [- 1, 1] \end{array}$

Question

Equation ax² + bx + c = 0 has

A

real and equal roots
B

complex roots
C

unequal positive real roots
D

unequal roots

Question

The solution set of $| | x + c | - 2 a | < 4 b$ is

A

[-2,2]
B

[0,2]
C

[-1,1]
D

[-2,1]

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