How do i find the limit of sin(3x)/sin(5x) as x->0 without applying L'hospital's rule?

just too easy, here's how

sin(3x)/sin(5x) =( sin(3x)/sin(5x) ) * ( 3/3 ) * ( 5/5 ) =

( 3/5 ) * ( (5 * sin(3x) )/(3 * sin (5x) ) ) = 3/5

as sin(3x)/3 =1 and 5/sin(5x) = 1 i.e. ( (5 * sin(3x) )/(3 * sin (5x) ) ) = 1/1 = 1

lim sin(x)/x = 1

Similarly lim sin(ax)/ax = 1

lim sin(3x)/sin(5x)

= lim [3*5x sin(3x)] / [5*3x sin(5x)]

= lim (3/5) [sin(3x)/(3x)] * [(5x)/sin(5x)]

= lim (3/5) [sin(3x)/(3x)] / [sin(5x)/(5x)]

= 3/5 * 1 / 1

= 3/5

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EDIT:

If you can reduce it to (3/5) sin(x)/sin(x), then you CAN cancel. This is NOT like dividing by zero, since we are dealing with LIMIT as x approaches 0.

This is a very common approach in limits. Manipulating the expression, so that you can cancel values that cause you to get a 0/0 result. This canceling of terms then removes this division of 0/0.

You should be aware of a well-known limit

sin(t)/t -> 1 as t->0 (it can be proved geometrically using a unit circle).

In our case,

sin(3x)/sin(5x)=

={sin(3x)/(3x)}/

/{sin(5x)/(5x)}*(3/5)

Denoting 3x=t, we'll get that t->0 as x->0. Similarly, if t=5x, then t->0 as x->0.

So the aforementioned limit applies and

lim (sin(3x)/sin(5x),x->0)=

=(3/5)*lim({sin(3x)/(3x)})/

/{lim(sin(5x)/(5x)})

=(3/5)*1/1=3/5

The order of importance of a logarithm is under the order of a linear function. So the numerator procedures 0 quicker than the denominator does. it is going to be 0. L'scientific institution's rule is incredibly evaluating the orders of strengthen of the function.

sin(x) / x = 1

Now, let's look at the problem

sin(3x) / sin(5x)

Multiply top and bottom by 3x leaving...

[sin(3x) / 3x] * [3x / sin(5x)]

[ 1 ] * [ 3x / sin(5x) ]

Multiply top and bottom by 5x leaving...

[ 5x / sin(5x) ] * [ 3x / 5x ]

[ 1 ] * (3/5)

3/5