Slove this limit please ?

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Answer 1 · 2017-03-18T14:45:01

# = 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#

Answer 2 · 2017-03-18T15:05:20

#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.#

Enjoy Maths.!