answer:Suppose that lim_{n to infty} sin (n^2) = l. Consider the subsequences b_{2k} = sin (4k^2) and b_{2k+1} = sin ((2k+1)^2). We have that lim_{k to infty} sin (4k^2) = 0 and lim_{k to infty} sin ((2k+1)^2) = 1. Since every subsequence of a convergent sequence must converge to the same limit, we have a contradiction. Therefore, the sequence b_n = sin (n^2) does not converge.

answer:We can see that the left side of the equation is a fraction involving the tangent and sine of 30 degrees, while the right side is the sine of 60 degrees. To determine if they are equivalent, we can simplify both sides using the trigonometric identities and exact values. After simplifying, we find that the left side is equal to {eq}dfrac{2sqrt{3}}{3} {/eq}, while the right side is equal to {eq}dfrac{sqrt{3}}{2} {/eq}. Since these two expressions are not equal, we can conclude that the left side of the equation is not equivalent to the right side. We're required to determine if the left side of the given trigonometric equation is equivalent to the right side, or not. dfrac{tan 30^circ}{sin 30^circ} = sin 60^circ As we refer to the table of exact values for each trigonometric expression and its special angles. We can substitute each value as, dfrac{dfrac{sqrt{3}}{3}}{dfrac{1}{2}} = dfrac{sqrt{3}}{2} Then, we apply the rules of dividing two fractions in the form, {eq}dfrac{dfrac{a}{b}}{dfrac{c}{d}} = left(dfrac{a}{b}right) left(dfrac{d}{c}right) {/eq} to simplify the left side of the equation begin{align} left(dfrac{sqrt{3}}{3}right) left(dfrac{2}{1}right) &= dfrac{sqrt{3}}{2} dfrac{2sqrt{3}}{3} &= dfrac{sqrt{3}}{2} end{align} Hence, from the given values of both sides of the equations, we can say that dfrac{2sqrt{3}}{3} ne dfrac{sqrt{3}}{2} Therefore, the left side is not equivalent to the right side of the equation.

answer:The addition of these two matrices is: [ left( begin{array}{ccc} 5 + (-3) & 9 + 1 & 1 + (-2) -7 + (-1) & 5 + (-7) & 0 + (-7) end{array} right) ] Which simplifies to: [ left( begin{array}{ccc} 2 & 10 & -1 -8 & -2 & -7 end{array} right) ]

answer:The function {eq}displaystyle lim_{x to 2} dfrac{3x^2 - 5x - 2}{x^2 - 4} {/eq} is only undefined at {eq}x^2 - 4 = 0 implies x = pm 2 {/eq}. Since {eq}2 = 2 {/eq}, we cannot apply direct substitution. Instead, we can factor the numerator and denominator: {eq}begin{align*} displaystyle lim_{x to 2} dfrac{3x^2 - 5x - 2}{x^2 - 4}& = lim_{x to 2} dfrac{(3x + 1)(x - 2)}{(x + 2)(x - 2)}& & = lim_{x to 2} dfrac{3x + 1}{x + 2}& & = dfrac{3(2) + 1}{2 + 2}& & =bf{frac{7}{4} } end{align*} {/eq}