How do you find the limit of # (sin 3x)/ (sin 4x)# as x approaches 0?

Question

Emily Ammerman · Accepted Answer

Use #lim_(thetararr0)sintheta/theta = 1# and some algebra.

Note: because #lim_(thetararr0)sintheta/theta = 1#, we also have #lim_(thetararr0)theta/sintheta = 1# Rewrite the expression to use the limits noted above. # (sin 3x)/ (sin 4x) = (sin3x)/1 * 1/(sin4x)# # = [(3x)/1 (sin3x)/(3x)]* [1/(4x) (4x)/(sin4x)]# # = (3x)/(4x)sin(3x)/(3x) (4x)/sin(4x)# #= 3/4 [(sin3x)/(3x) (4x)/(sin4x)]# Now, as #xrarr0#, #(3x)rarr0# so #(sin3x)/(3x) rarr1#. (Using #theta = 3x#) And, as #xrarr0#, #(4x)rarr0# so #(4x)/(sin4x) rarr1#. (Using #theta = 4x#) Therefore the limit is #3/4#. #lim_(xrarr0)(sin3x)/(sin4x) = lim_(xrarr0)3/4 (sin3x)/(3x)(4x)/(sin4x)# # = 3/4 lim_(xrarr0)(sin3x)/(3x)lim_(xrarr0)(4x)/(sin4x)# # = 3/4(1)(1) = 3/4#

William Abdalla · Answer

undefined To find the limit of (sin 3x) / (sin 4x) as x approaches 0, we can use the concept of L'Hôpital's Rule. By applying this rule, we differentiate the numerator and denominator separately with respect to x.

Differentiating the numerator, we get: d/dx (sin 3x) = 3cos 3x.

Differentiating the denominator, we get: d/dx (sin 4x) = 4cos 4x.

Now, we can evaluate the limit by substituting x = 0 into the derivatives we obtained:

lim (x→0) (3cos 3x) / (4cos 4x).

Plugging in x = 0, we have:

(3cos 0) / (4cos 0) = 3/4.

Therefore, the limit of (sin 3x) / (sin 4x) as x approaches 0 is 3/4.