Question

Determine lim x->0 sin5x / 2x

Answer 1

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Answer 2

Answer:

The value of the given limit is

$\lim_{x \to 0} \frac{\sin5x}{2x}=\frac{5}{2}$

Step-by-step explanation:

The given limit is

$\lim_{x \to 0} \frac{\sin5x}{2x}$

By substituting the value of the limit we get

$\lim_{x \to 0} \frac{\sin5x}{2x}=\frac{\sin0}{0} =\frac{0}{0}$

This is an indeterminate form and thus to solve it we shall use the L-hospital rule in which both the numerator and the denominator of the limit are differentiated until a finite value is obtained.

Applying the L-hospital rule we get,

$\lim_{x \to 0} \frac{\frac{d}{dx}(\sin5x)}{\frac{d}{dx}(2x)}=\lim_{x \to 0} \frac{5\cos5x}{2}$

By substituting the value of the limit we get

$\lim_{x \to 0} \frac{5\cos5x}{2}=\frac{5\cos0}{2} =\frac{5}{2}$

Therefore,

The value of the given limit is

$\lim_{x \to 0} \frac{\sin5x}{2x}=\frac{5}{2}$

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