First slide

Introduction to limits

Question

$If \lim_{x \to 0} \frac{f (x)}{\sin^{2} x} = 8, \lim_{x \to 0} \frac{g (x)}{2 cosx - {xe}^{x} + x^{3} + x - 2} = λ and \lim_{x \to 0} {(1 + 2 f (x))}^{\frac{1}{g (x)}} = \frac{1}{e}, then$

Moderate

Question

The value of $λ$ is

A

2
B

4
C

8
D

16

Solution

$\begin{array}{l} \lim_{x \to 0} \frac{f (x)}{\sin^{2} x} = 8 \\ \Rightarrow \lim_{x \to 0} \frac{\frac{f (x)}{x^{2}}}{\frac{\sin^{2} x}{x^{2}}} = 8 \\ \Rightarrow \lim_{x \to 0} \frac{f (x)}{x^{2}} = 8 . . . (1) \\ Also, \lim_{x \to 0} \frac{g (x)}{2 cosx - {xe}^{x} + x^{3} + x - 2} = λ \\ \Rightarrow \lim_{x \to 0} \frac{\frac{g (x)}{x^{2}}}{\frac{- 2 (1 - cosx)}{x^{2}} - \frac{e^{x} - 1}{x} + x} = λ \\ \Rightarrow \lim_{x \to 0} \frac{\frac{g (x)}{x^{2}}}{\frac{- 4 \sin^{2} \frac{x}{2}}{x^{2}} - 1 + 0} = λ \\ \Rightarrow \lim_{x \to 0} \frac{g (x)}{x^{2}} = - 2 λ . . . (2) \\ Now, \lim_{x \to 0} {(1 + 2 f (x))}^{\frac{1}{g (x)}} (one power infinity form) \\ = e^{\lim_{x \to 0} \frac{2 f (x)}{g (x)}} = e^{\lim_{x \to 0} \frac{2 (f (x) / x^{2})}{(g (x) / x^{2})}} \\ = e^{\lim_{x \to 0} \frac{16}{- 2 λ}} \\ = e^{- \frac{8}{λ}} \\ = e^{- 1} (given) \\ ∴ λ = 8 \\ \lim_{x \to 0} {(1 + f (x))}^{\frac{1}{2 g (x)}} = e^{\lim_{x \to 0} \frac{f (x)}{2 g (x)}} \\ = e^{\lim_{x \to 0} \frac{(f (x) / x^{2})}{2 (g (x) / x^{2})}} \\ = e^{\lim_{x \to 0} \frac{8}{2 \times (- 16)}} \\ = e^{- \frac{1}{4}} \end{array}$

Question

$\lim_{x \to 0} {(1 + f (x))}^{\frac{1}{2 g (x)}}$ is equal to

A

$e^{- \frac{1}{4}}$
B

$e^{- \frac{1}{2}}$
C

$e^{- 2}$
D

$e^{- 4}$

Solution

$\begin{array}{l} \lim_{x \to 0} \frac{f (x)}{\sin^{2} x} = 8 \\ \Rightarrow \lim_{x \to 0} \frac{\frac{f (x)}{x^{2}}}{\frac{\sin^{2} x}{x^{2}}} = 8 \\ \Rightarrow \lim_{x \to 0} \frac{f (x)}{x^{2}} = 8 . . . (1) \\ Also, \lim_{x \to 0} \frac{g (x)}{2 cosx - {xe}^{x} + x^{3} + x - 2} = λ \\ \Rightarrow \lim_{x \to 0} \frac{\frac{g (x)}{x^{2}}}{\frac{- 2 (1 - cosx)}{x^{2}} - \frac{e^{x} - 1}{x} + x} = λ \\ \Rightarrow \lim_{x \to 0} \frac{\frac{g (x)}{x^{2}}}{\frac{- 4 \sin^{2} \frac{x}{2}}{x^{2}} - 1 + 0} \\ \Rightarrow \lim_{x \to 0} \frac{g (x)}{x^{2}} = - 2 λ . . . (2) \\ Now, \lim_{x \to 0} {(1 + 2 f (x))}^{\frac{1}{g (x)}} (one power infinity form) \\ = e^{\lim_{x \to 0} \frac{2 f (x)}{g (x)}} = e^{\lim_{x \to 0} \frac{2 (f (x) / x^{2})}{(g (x) / x^{2})}} \\ = e^{\lim_{x \to 0} \frac{16}{- 2 λ}} \\ = e^{- \frac{8}{λ}} \\ = e^{- 1} (given) \\ ∴ λ = 8 \\ \lim_{x \to 0} {(1 + f (x))}^{\frac{1}{2 g (x)}} = e^{\lim_{x \to 0} \frac{f (x)}{2 g (x)}} \\ = e^{\lim_{x \to 0} \frac{(f (x) / x^{2})}{2 (g (x) / x^{2})}} \\ = e^{\lim_{x \to 0} \frac{8}{2 \times (- 16)}} \\ = e^{- \frac{1}{4}} \end{array}$

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