Evaluate: limx→128x-32x-1-4x2+14x2-1 - Mathematics

Question

Evaluate: $lim_{x \to \frac{1}{2}} \frac{8 x - 3}{2 x - 1} - \frac{4 x^{2} + 1}{4 x^{2} - 1}$

Sum

Solution

Given that $lim_{x \to \frac{1}{2}} (\frac{8 x - 3}{2 x - 1} - \frac{4 x^{2} + 1}{4 x^{2} - 1})$

= $lim_{x \to \frac{1}{2}} [\frac{(8 x - 3) (2 x + 1) - (4 x^{2} + 1)}{(4 x^{2} - 1)}]$

= $lim_{x \to \frac{1}{2}} [\frac{16 x^{2} - 6 x + 8 x - 3 - 4 x^{2} - 1}{4 x^{2} - 1}]$

= $lim_{x \to \frac{1}{2}} [\frac{12 x^{2} + 2 x - 4}{4 x^{2} - 1}]$

= $lim_{x \to \frac{1}{2}} \frac{2 (6 x^{2} + x - 2)}{4 x^{2} - 1}$

= $lim_{x \to \frac{1}{2}} \frac{2 [6 x^{2} + 4 x - 3 x - 2]}{(2 x + 1) (2 x - 1)}$

= $lim_{x \to \frac{1}{2}} \frac{2 [2 x (3 x + 2) - 1 (3 x + 2)]}{(2 x + 1) (2 x - 1)}$

= $lim_{x \to \frac{1}{2}} \frac{2 (3 x + 2) (2 x - 1)}{(2 x + 1) (2 x - 1)}$

= $lim_{x \to \frac{1}{2}} \frac{2 (3 x + 2)}{(2 x + 1)}$

Taking limit, we have

= $\frac{2 (3 \times \frac{1}{2} + 2)}{2 \times \frac{1}{2} + 1}$

= $\frac{2 (\frac{7}{2})}{2}$

= $\frac{7}{2}$

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Limits of Trigonometric Functions

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Chapter 13: Limits and Derivatives - Exercise [Page 240]

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Evaluate: limx→128x-32x-1-4x2+14x2-1 - Mathematics

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