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

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Answer 1 · 2016-04-24T14:18:08

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#