Slove this limit please ?

Answer 1

# = 1/5#

We use well known trig limit:

# lim_(theta to 0) (sin theta)/(theta) = 1 #

Or, specifically, it's inverse:

# lim_(theta to 0) theta/(sin theta) = 1 #

So:

# lim_(x to 0) (5x^2)/(sin^2 5x)#
#1/5 lim_(x to 0) (5x)/(sin 5x) cdot (5x)/(sin 5x) #

And by the product rule for limits:

#= 1/5 lim_(x to 0) (5x)/(sin 5x) cdot lim_(x to 0) (5x)/(sin 5x) #
If we say that #theta = 5x#, we have:
#= 1/5 lim_(theta to 0) (theta)/(sin theta) cdot lim_(x to 0) (theta)/(sin theta) #
#= 1/5 cdot 1 cdot 1 = 1/5#
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Answer 2

#1/5.#

Recall that, #lim_(theta to 0) theta/sintheta=1.#
Now, the Reqd. Lim.=#lim_(x to 0)(5x^2)/(sin^2(5x))#
#=lim_(x to 0)5{x/sin(5x)}^2#
#=lim_(x to 0)5{(5x)/sin(5x)*1/5}^2.#
#=5(1/5)^2{lim_((5x) to 0)(5x)/sin(5x)}^2.#
#=1/5(1)^2.#
#=1/5.#

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Answer 3

Certainly! Please provide the limit you would like me to solve for you.

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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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