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

Answer 1

#1/4.#

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.#
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Answer 2

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

Applying the triangular 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 #
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Answer 3

#1/4#

The first step in solving this limit is to use direct substitution, which has the following appearance.

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

Since the answer is indeterminate, we would apply L'Hôpital's rule, which states that the numerator and denominator should be treated as independent functions and that their respective derivatives should be taken as follows:

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

In this instance:

#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#

Your limit exists.

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Answer 4

To evaluate the limit lim_(x→0) sin(x)/sin(4x), we can use L'Hôpital's Rule, which states that if the limit of the ratio of two functions exists as x approaches a certain value, and if both functions are differentiable and the limit of their derivatives also exists, then the limit of the ratio is the same as the limit of the ratio of their derivatives.

Applying L'Hôpital's Rule:

lim_(x→0) (sin(x)/sin(4x)) = lim_(x→0) (cos(x)/4cos(4x))

Now, evaluate the limit as x approaches 0:

lim_(x→0) (cos(x)/4cos(4x)) = cos(0)/(4cos(0)) = 1/4

So, the limit lim_(x→0) sin(x)/sin(4x) equals 1/4.

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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.

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.

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.

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.

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