limit as x approaches infinity. (sin2x)/(x)
answer is 0. show me how please. thanks in advance =)
As x approaches infinity, the range of sin(2x) takes on the upper and lower limits of +1 and -1.
Since Lim 1/∞ and Lim -1/∞ both equal to zero as x→∞, we conclude by the "sandwich rule" or the "squeeze theorem" that the limit of sin(2x)/x lies between the two, namely 0.
To find the limit as x approaches infinity for the expression (sin(2x))/(x), we can use a technique called L'Hospital's Rule or the rule of De L'Hôpital.
First, let's take the derivative of the numerator and the denominator separately. The derivative of sin(2x) with respect to x is 2cos(2x), and the derivative of x with respect to x is 1.
Now, we have (2cos(2x))/(1) as our new expression.
Next, we can evaluate the limit as x approaches infinity for this new expression. Since cos(2x) oscillates between -1 and 1, and our denominator is a constant 1, we notice that the numerator will fluctuate between -2 and 2 as x approaches infinity.
Therefore, as x approaches infinity, the limit of (2cos(2x))/(1) will oscillate, but it will always be between -2 and 2.
Hence, the limit of (sin(2x))/(x) as x approaches infinity is 0.