How do you find #lim sin(2x)/x# as #x->0# using l'Hospital's Rule?

Answer 1

Charles Anctil

#lim_(x->0) sin(2x)/x = 2#

The limit:

#lim_(x->0) sin(2x)/x# is in the indeterminate form #0/0# so we can solve it using l'Hospital's rule:

#lim_(x->0) sin(2x)/x = lim_(x->0) (d/dx sin(2x))/(d/dx x) = lim_(x->0) (2cos(2x))/1 = 2#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Cameron Alexander

To find ( \lim_{{x \to 0}} \frac{{\sin(2x)}}{x} ) using l'Hopital's Rule, we can differentiate the numerator and denominator separately with respect to ( x ) until we can evaluate the limit directly:

( \lim_{{x \to 0}} \frac{{\sin(2x)}}{x} )

Applying l'Hopital's Rule once:

( = \lim_{{x \to 0}} \frac{{2\cos(2x)}}{1} )

Evaluating at ( x = 0 ), we get:

( = 2\cos(0) = 2 )

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

How do you find #lim sin(2x)/x# as #x-&gt;0# using l'Hospital's Rule?

How do you find #lim sin(2x)/x# as #x->0# using l'Hospital's Rule?