limit x tends to infinitive ((e^x)-1)/(x)
(2)-1/3 x (4)-1/3 = (2 x 4)-1/3 = (8)-1/3 = 1/81/3 = 1/2
Any number x raised to a negative power -y is equivalent to the reciprocal of x raised to y. So, 10-73 would be the fraction 1 over the number 1 with 73 zeroes after it. A very small number indeed!
Yes, ∞ and -∞ both exist as distinct entities. If I take the limit of 1/x as x approaches 0 from the positive side, I get ∞. On the other hand if I take the limit of 1/x as x approaches 0 from the negative side, I get -∞.
1
limit x tends to infinitive ((e^x)-1)/(x)
(2)-1/3 x (4)-1/3 = (2 x 4)-1/3 = (8)-1/3 = 1/81/3 = 1/2
It is always negative when raised to an odd power and always positive when raised to an even power -2 to the third power = -2 x -2 x -2 = -8 -2 to the fourth power = -2 x -2 x -2 x -2 = +16
Any number x raised to a negative power -y is equivalent to the reciprocal of x raised to y. So, 10-73 would be the fraction 1 over the number 1 with 73 zeroes after it. A very small number indeed!
Yes, ∞ and -∞ both exist as distinct entities. If I take the limit of 1/x as x approaches 0 from the positive side, I get ∞. On the other hand if I take the limit of 1/x as x approaches 0 from the negative side, I get -∞.
1
The limit is 4.
The answer to -x^5y^2 is a negative number raised to the fifth power multiplied by y raised to the second power. It cannot be simplified further without knowing the specific values of x and y.
a negative number plus a negative number is negative. here is a proof.(-x)+(-x)(-1)x+(-1)x-1(x+x)-1(2x)-2xa negative number times a negative number is positive though.
The limit of cos2(x)/x as x approaches 0 does not exist. As x approaches 0 from the left, the limit is negative infinity. As x approaches 0 from the right, the limit is positive infinity. These two values would have to be equal for a limit to exist.
The negative reciprocal of a non-zero number x, is -1/x. And the negative reciprocal of that is -1/(-1/x) = -1*(x/-1) = x
There in so limit. If x is any positive number, then (x + 1) is further from 0.If y is any negative number, then (y - 1) is further from 0.