First slide

Introduction to limits

Question

$If f^{1} (0) = 3, then lim_{x \to 0} \frac{x^{2}}{f (x^{2}) - 6 f (4 x^{2}) + 5 f (7 x^{2})} =$

Moderate

A

$\frac{1}{36}$
B

$- \frac{1}{36}$
C

$\frac{1}{34}$
D

$\frac{1}{106}$

Solution

$\begin{array}{l} By L-Hospital rule \\ lim_{x \to 0} \frac{2 x}{f^{1} (x^{2}) (2 x) - 6 f^{1} (4 x^{2}) (8 x) + 5 f^{1} (7 x^{2}) 14 x} \\ = lim_{x \to 0} \frac{2}{f^{1} (x^{2}) 2 - 48 f^{1} (4 x^{2}) + 70 f^{1} (7 x^{2})} \\ = \frac{2}{2 f^{1} (0) - 48 f^{1} (0) + 70 f^{1} (0)} \\ = \frac{2}{6 - 48 (3) + 70 (3)} \\ = \frac{2}{3 [2 - 48 + 70]} \end{array}$
$= \frac{1}{36}$

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