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.!