How can the limit as x approaches 0 of (sin x) /x =1?
There is a Taylor expansion for such functionssin(x) = x - (x^3)/3! + (x^5)/5! +…+ (-1)^3 * (x^(2n+1))/(2n+1) + o(x^(2n+1) where n>=0The degree of the function in the denominator is 1. So, take n to be 1Then, sin(x) becomes:sin(x) = x + o(x)So, sin(x)/x = 1 + o(1).Finally, lim(x->0) sin(x)/x = 1 ( simple).Note: Please, try to understand the scene behind the expansion of Taylor. If you can use it comfortably, you can calculate difficult limits easily :)))
Is it right to say that the the limit of sin(x) as x approaches 0 is x?
Please note that [math]\lim_{x \to 0} \sin(x) = 0[/math]The right way to put it across is for small values of [math]x, \sin(x) \approx x[/math]How small, the value of [math]x[/math] can be?Consider the angle [math]10^o = 0.174533 \text{ rad}[/math][math]\sin(10^o) = \sin(0.174533 \text{ rad}) = 0.173648 [/math]If approximation of [math]\sin(x) = x[/math] is applied for [math]10^o[/math], margin of error = [math]0.5\text{%}[/math]For all values of [math]|x| \lt 10^o \text{ OR } |x| \lt 0.174533 \text{ rad},[/math] one can say that [math]\sin x \approx x[/math]NOTE: [math]x [/math] must be expressed in radians for this approximation
Lim. as x approaches 0 of ( (√1+tanx) - (√1+sinx) ) / x^3?
Multiply top and bottom by (sqrt(1 + tan(x)) + sqrt(1 + sin(x))) to get lim(x->0)[(1 + tan(x) - (1 + sin(x))) / (x^3 * (sqrt(1 + tan(x)) + sqrt(1 + sin(x))))] = lim(x->0)[((sin(x)/cos(x)) - sin(x)) / (x^3 * (sqrt(1 + tan(x)) + sqrt(1 + sin(x))))] = lim(x->0)[(sin(x) / x)*(1/cos(x))*((1 - cos(x))/x^2) / (sqrt(1+tan(x)) + sqrt(1+sin(x)))] = 1*1*(1/2)*lim(x->0)[((1 - cos(x)) / x^2) * ((1 + cos(x))/(1 + cos(x))] = (1/2)*lim(x->0)(1/(1 + cos(x)))*lim(x->0)(sin^2(x) / x^2) = (1/2)*(1/2)*[lim(x->0)(sin(x)/x)]^2 = 1/4. I condensed the last few steps a bit; I hope you can still follow the process. :)
Limit of 1 - sin x OVER cos x when x approaches pi/2?
lim (1 - sinx) / cosx = indefinite form (0/0) x → π/2 multiply and divide by (1 + sinx) as: lim [(1 - sinx)(1 + sinx)] / [cosx (1 + sinx)] = x → π/2 expand the numerator into a difference of squares: lim (1 - sin²x) / [cosx (1 + sinx)] = x → π/2 replace (1 - sin²x) with cos²x: lim cos²x / [cosx (1 + sinx)] = x → π/2 cancel (cosx) out: lim cosx / (1 + sinx) = 0 /(1 + 1) = 0 x → π/2 I hope it helps.. Bye!
As the x approaches the limit of 4 of (3+x^1/2) evaluate the limit if possible?
If you mean lim (3+x^1/2) as x->4 then this is 3+sqrt4=5
What is the answer of limit x tends to infinity sqrt ((x - sinx) / (x + cos^2x)?
[math]L = \lim \limits_{x\to \infty} \sqrt{\frac{x-\sin x}{x+ \cos^2 x}}.[/math]Let [math]x=\frac{1}{y} \qquad \Rightarrow \qquad[/math] As [math]x\to \infty, \qquad y\to 0.[/math][math]\Rightarrow \qquad L = \lim \limits_{y\to 0} \sqrt{\frac{\frac{1}{y}-\sin \left(\frac{1}{y}\right)}{\frac{1}{y}+ \cos^2 \left(\frac{1}{y}\right)}} = \lim \limits_{y\to 0} \sqrt{\frac{1-y\sin \left(\frac{1}{y}\right)}{1+ y\cos^2 \left(\frac{1}{y}\right)}}.[/math][math]\sin \left(\frac{1}{y}\right)[/math] and [math]\cos^2 \left(\frac{1}{y}\right)[/math] are both bounded functions.[math]\sin \left(\frac{1}{y}\right) \,\,\in \,\, [-1,1] \,\, \forall \,\,y\,\,\in \,\,R[/math] and [math]\cos^2 \left(\frac{1}{y}\right) \,\,\in \,\, [0,1] \,\, \forall \,\,y\,\,\in \,\,R.[/math][math]\Rightarrow \qquad \lim \limits_{y\to 0} y\sin \left(\frac{1}{y}\right)=0[/math] and [math]\lim \limits_{y\to 0} y\cos^2 \left(\frac{1}{y}\right)=0.[/math][math]\Rightarrow \qquad \lim \limits_{y\to 0} \sqrt{\frac{1-y\sin \left(\frac{1}{y}\right)}{1+ y\cos^2 \left(\frac{1}{y}\right)}}=\lim \limits_{y\to 0} \sqrt{\frac{1-0}{1+ 0}} = 1.[/math][math]\Rightarrow \qquad \lim \limits_{x\to \infty} \sqrt{\frac{x-\sin x}{x+ \cos^2 x}}=1.[/math]