Question

Evaluate the limit `lim_(x rarr 0) sin(x)/sin(4x)`?

Answer 1

`1/4.`

Explanation:

We will use the Standard Limit `lim_(thetato0)sintheta/theta=1.`

Now, the Reqd. Lim.`=lim_(xto0)sinx/sin(4x)`

`=lim_(xto0){(sinx/x)(x)}/{(sin(4x))/(4x)(4x)}`

`=(1/4){lim_(xto0)sinx/x}/{lim_((4x)to0)(sin(4x))/(4x)}`

`=1/4{1/1}`

`=1/4.`

Answer 2

`1/4`

Explanation:

When solving this limit, the first step is to use direct substitution which looks like this

`lim x->0 (sin(x))/(sin(4x)) = (sin(0))/(sin(0)) = 0/0`

This answer is indeterminate and therefore we would use L'Hôpital's rule which says to treat the numerator and denominator as separate functions and to take the derivative of each one like this

L'Hôpital's Rule: `f(x)/g(x) = 0/0 = (f'(x))/(g'(x))`

This case:

`f(x)=sin(x) g(x)=sin(4x)`
`f'(x)=cos(x) g'(x)=4cos(4x)`

`(f'(x))/(g'(x)) = (cos(x))/(4cos(4x))`
Use the same limit
`lim x->0 (cos(x))/(4cos(4x))`
Substitute the value of the limit into the new expression

`(cos(0))/(4cos(0)) = 1/4`

There is your limit

Answer 3

`lim_(x rarr 0) sinx/sin(4x) = 1/4`

Explanation:

Using the trig identity:

`sin 2A -= 2sinAcosA`

We have:

`lim_(x rarr 0) sinx/sin(4x) = lim_(x rarr 0) sinx/sin(2(2x))`
`" " = lim_(x rarr 0) sinx/(2sin2xcos2x)`
`" " = lim_(x rarr 0) sinx/(2(2sinxcosx)cos2x)`
`" " = lim_(x rarr 0) sinx/(4sinxcosxcos2x)`
`" " = lim_(x rarr 0) 1/(4cosxcos2x)`
`" " = 1/(4*1*1)`
`" " = 1/4`