First slide

Introduction to limits

Question

If $L = \lim_{x \to 0} \frac{sinx + {ae}^{x} + {be}^{- x} + c \log_{e} (1 + x)}{x^{3}}$ exists finitely, then

Moderate

Question

The value of L is

A

$\frac{1}{2}$
B

$- \frac{1}{3}$
C

$- \frac{1}{6}$
D

3

Solution

$\begin{array}{l} L = \lim_{x \to 0} \frac{sinx + {ae}^{x} + {be}^{- x} + c \log_{e} (1 + x)}{x^{3}} \\ = \lim_{x \to 0} \frac{(x - \frac{x^{3}}{3!}) + a (1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \frac{x^{3}}{3!}) + b (1 - \frac{x}{1!} + \frac{x^{2}}{2!} - \frac{x^{3}}{3!}) + c (x - \frac{x^{2}}{2} + \frac{x^{3}}{3})}{x^{3}} \\ = \lim_{x \to 0} \frac{(a + b) + (1 + a - b + c) + (\frac{a}{2} + \frac{b}{2} - \frac{c}{2}) x^{2} + (- \frac{1}{3!} + \frac{a}{3!} - \frac{b}{3!} + \frac{c}{3}) x^{3}}{x^{3}} \\ \Rightarrow a + b = 0, 1 + a - b + c = 0, \frac{a}{2} + \frac{b}{2} - \frac{c}{2} = 0 \\ and L = - \frac{1}{3!} + \frac{a}{3!} - \frac{b}{3!} + \frac{c}{3} \\ Solving first three equations we get c = 0, a = - \frac{1}{2}, b = \frac{1}{2} . \\ ∴ L = \frac{1}{3} \\ Equation {ax}^{2} + bx + c = 0 reduces to x^{2} - x = 0 \Rightarrow x = 0, 1 \\ ||x + c| - 2 a| < 4 b reduces to |x| + 1 < 2 \\ \Rightarrow - 2 < |x| + 1 < 2 \\ \Rightarrow 0 \leq |x| < 1 \\ \Rightarrow x \in [- 1, 1] \end{array}$

Question

Equation ${ax}^{2} + bx + c = 0$ has

A

real and equal roots
B

complex roots
C

unequal positive real roots
D

unequal roots

Solution

$\begin{array}{l} L = \lim_{x \to 0} \frac{sinx + {ae}^{x} + {be}^{- x} + c \log_{e} (1 + x)}{x^{3}} \\ = \lim_{x \to 0} \frac{(x - \frac{x^{3}}{3!}) + a (1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \frac{x^{3}}{3!}) + b (1 - \frac{x}{1!} + \frac{x^{2}}{2!} - \frac{x^{3}}{3!}) + c (x - \frac{x^{2}}{2} + \frac{x^{3}}{3})}{x^{3}} \\ = \lim_{x \to 0} \frac{(a + b) + (1 + a - b + c) + (\frac{a}{2} + \frac{b}{2} - \frac{c}{2}) x^{2} + (- \frac{1}{3!} + \frac{a}{3!} - \frac{b}{3!} + \frac{c}{3}) x^{3}}{x^{3}} \\ \Rightarrow a + b = 0, 1 + a - b + c = 0, \frac{a}{2} + \frac{b}{2} - \frac{c}{2} = 0 \\ and L = - \frac{1}{3!} + \frac{a}{3!} - \frac{b}{3!} + \frac{c}{3} \\ Solving first three equations we get c = 0, a = - \frac{1}{2}, b = \frac{1}{2} . \\ ∴ L = \frac{1}{3} \\ Equation {ax}^{2} + bx + c = 0 reduces to x^{2} - x = 0 \Rightarrow x = 0, 1 \\ ||x + c| - 2 a| < 4 b reduces to |x| + 1 < 2 \\ \Rightarrow - 2 < |x| + 1 < 2 \\ \Rightarrow 0 \leq |x| < 1 \\ \Rightarrow x \in [- 1, 1] \end{array}$

Question

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

A

$[- 2, 2]$
B

$[0, 2]$
C

$[- 1, 1]$
D

$[- 2, 1]$

Solution

$\begin{array}{l} L = \lim_{x \to 0} \frac{sinx + {ae}^{x} + {be}^{- x} + c \log_{e} (1 + x)}{x^{3}} \\ = \lim_{x \to 0} \frac{(x - \frac{x^{3}}{3!}) + a (1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \frac{x^{3}}{3!}) + b (1 - \frac{x}{1!} + \frac{x^{2}}{2!} - \frac{x^{3}}{3!}) + c (x - \frac{x^{2}}{2} + \frac{x^{3}}{3})}{x^{3}} \\ = \lim_{x \to 0} \frac{(a + b) + (1 + a - b + c) + (\frac{a}{2} + \frac{b}{2} - \frac{c}{2}) x^{2} + (- \frac{1}{3!} + \frac{a}{3!} - \frac{b}{3!} + \frac{c}{3}) x^{3}}{x^{3}} \\ \Rightarrow a + b = 0, 1 + a - b + c = 0, \frac{a}{2} + \frac{b}{2} - \frac{c}{2} = 0 \\ and L = - \frac{1}{3!} + \frac{a}{3!} - \frac{b}{3!} + \frac{c}{3} \\ Solving first three equations we get c = 0, a = - \frac{1}{2}, b = \frac{1}{2} . \\ ∴ L = \frac{1}{3} \\ Equation {ax}^{2} + bx + c = 0 reduces to x^{2} - x = 0 \Rightarrow x = 0, 1 \\ ||x + c| - 2 a| < 4 b reduces to |x| + 1 < 2 \\ \Rightarrow - 2 < |x| + 1 < 2 \\ \Rightarrow 0 \leq |x| < 1 \\ \Rightarrow x \in [- 1, 1] \end{array}$

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