First slide

Introduction to limits

Question

Letf(x) be a polynomial satisfying $\lim_{x \to \infty} \frac{x^{2} f (x)}{2 x^{5} + 3} = 6$ and $f (1) = 3, f (3) = 7 and f (5) = 11 .$ then

Moderate

Question

The value of f(0) is

A

120
B

119
C

180
D

-179

Solution

$\begin{array}{l} Since \lim_{x \to \infty} \frac{x^{2} f (x)}{2 x^{5} + 3} = 6, which is finite non - zero, f (x) must be polynomial of degree' 3' . \\ Also f (1) = 3, f (3) = 7 and f (5) = 11 \\ ∴ f (x) = λ (x - 1) (x - 3) (x - 5) + (2 x + 1) \\ \Rightarrow \lim_{x \to \infty} \frac{x^{2} (λ (x - 1) (x - 3) (x - 5) + 2 x + 1)}{2 x^{5} + 3} = 6 \\ \Rightarrow \frac{λ}{2} = 6 \\ \Rightarrow λ = 12 \\ ∴ f (x) = 12 (x - 1) (x - 3) (x - 5) + (2 x + 1) \\ So, f (0) = 12 (- 1) (- 3) (- 5) + 1 = - 179 \\ \lim_{x \to 1} \frac{x - 1}{\sin (f (x) - 2 x - 1)} \\ = \lim_{x \to 1} \frac{x - 1}{\sin [12 (x - 1) (x - 3) (x - 5)]} \\ = \lim_{x \to 1} \frac{12 (x - 1) (x - 3) (x - 5)}{\sin [12 (x - 1) (x - 3) (x - 5)]} . \frac{1}{12 (x - 3) (x - 5)} \\ = \frac{1}{96} \end{array}$

Question

$\lim_{x \to 1} \frac{x - 1}{\sin (f (x) - 2 x - 1)}$ is equal to

A

$\frac{1}{96}$
B

$\frac{1}{48}$
C

$\frac{1}{24}$
D

1

Solution

$\begin{array}{l} Since \lim_{x \to \infty} \frac{x^{2} f (x)}{2 x^{5} + 3} = 6, which is finite non - zero, f (x) must be polynomial of degree' 3' . \\ Also f (1) = 3, f (3) = 7 and f (5) = 11 \\ ∴ f (x) = λ (x - 1) (x - 3) (x - 5) + (2 x + 1) \\ \Rightarrow \lim_{x \to \infty} \frac{x^{2} (λ (x - 1) (x - 3) (x - 5) + 2 x + 1)}{2 x^{5} + 3} = 6 \\ \Rightarrow \frac{λ}{2} = 6 \\ \Rightarrow λ = 12 \\ ∴ f (x) = 12 (x - 1) (x - 3) (x - 5) + (2 x + 1) \\ So, f (0) = 12 (- 1) (- 3) (- 5) + 1 = - 179 \\ \lim_{x \to 1} \frac{x - 1}{\sin (f (x) - 2 x - 1)} \\ = \lim_{x \to 1} \frac{x - 1}{\sin [12 (x - 1) (x - 3) (x - 5)]} \\ = \lim_{x \to 1} \frac{12 (x - 1) (x - 3) (x - 5)}{\sin [12 (x - 1) (x - 3) (x - 5)]} . \frac{1}{12 (x - 3) (x - 5)} \\ = \frac{1}{96} \end{array}$

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