# How do you do lim(x->0) sin7x/x and lim(x->0) tanx/4x?

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• Anonymous

We can use the fact that:

lim (x-->0) sin(x)/x = lim (x-->0) sin(ax)/(ax) = 1.

Then, we have:

lim (x-->0) sin(7x)/x

= lim (x-->0) [7 * sin(7x)/(7x)], by multiplying and dividing by 7

= 7 * lim (x-->0) sin(7x)/(7x)

= 7 * 1, from the above special limit

= 7.

For the second limit, we can use tan(x) = sin(x)/cos(x) to get:

lim (x-->0) tan(x)/(4x)

= lim (x-->0) [sin(x)/cos(x)]/(4x)

= lim (x-->0) sin(x)/[4x*cos(x)]

= lim (x-->0) [sin(x)/(4x) * 1/cos(x)]

= lim (x-->0) sin(x)/(4x) * lim (x-->0) 1/cos(x)

= 1/4 * lim (x-->0) sin(x)/x * lim (x-->0) 1/cos(x)

= 1/4 * 1 * 1/1

= 1/4.

I hope this helps!

• 5 years ago

have you ever discovered spinoff? If particular, then we could hire the l'Hôpital's rule: lim x->0 (f(x)/g(x)) = lim x->0 (f'(x)/g'(x)) = lim x->0 (f"(x)/g"(x)) =... etc. (be conscious heavily that the remarkable-term function and the backside-term function are derived independently) lim x->0 (6x^5)/(6sinx-6x+x^3) --> For the l'Hôpital's rule: f(x) = 6x^5 and g(x) = 6sinx-6x+x^3 1st spinoff --> lim x->0 (30x^4)/(6cosx-6+3x^2) = 0/0 2d spinoff --> lim x->0 (120x^3)/(-6sinx+6x) = 0/0 third spinoff --> lim x->0 (360x^2)/(-6cosx+6) = 0/0 4th spinoff --> lim x->0 (720x)/(6sinx) = 0/0 5th spinoff --> lim x->0 (720)/(6cosx) =720/6 =one hundred twenty So, the respond is one hundred twenty (THE decrease DOES EXIST !!!). NOPE, the function f(x) = (6x^5)/((6sinx)-6x+x^3) isn't oscillatory discontinuous at x=0. WHY? because of the fact: a million. we could make f(x) exist at x=0 by potential of including a factor for x=0 2. The function is non-give up for all x different than 0 ===> In different be conscious, f(x) has detachable discontinuity at x=0 hint: seem on the graph made by potential of Wolfram web site (source no. 2)