# Find the limit: `lim_(x->0) (sin (sqrt (2x))/(sqrt (2x)))`

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Tushar Chandra | Certified Educator

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Let's use simple symbols to make things easier. You want the value of :

`lim_(x->0) (sin (sqrt (2x))/(sqrt (2x)))` .

Now if you substitute the value of x = 0 in the expression we are finding the limit of, you get (sin 0)/0 = 0/0, which is indeterminate.

In these cases, an easy way to find the limit is by using l'Hopital's rule, which allows us to substitute the numerator and denominator with the derivative.

The derivative of `sin(sqrt(2x))` isĀ `2cos(sqrt 2x)*(1/2)/sqrt(2x)` = `cos(sqrt(2x))/sqrt(2x)`

Similarly the derivative of `sqrt(2x)` is `2(1/2)/sqrt(2x)` = `1/sqrt(2x)`

This gives the limit as : `lim_(x->0)((cos sqrt(2x)/sqrt(2x))/(1/sqrt(2x)))` = `lim_(x->0) cos sqrt(2x)`

Now if you substitute x = 0, you get cos 0 which is equal to 1.

The required value of `lim_(x->0) (sin (sqrt (2x))/(sqrt (2x)))` = 1.

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