lim x -> -inf [x/ex] = lim x -> +inf[-x/e-x] = - lim x -> +inf [ xex ] = -inf
If you want to see this function then I suggest you use either:
(a) wolframalpha.com: put in show me x/exp(x)
or (b) geogebra, which is available for the desktop.
The limit does not exist.
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.
1
When the limit as the function approaches from the left, doesn't equal the limit as the function approaches from the right. For example, let's look at the function 1/x as x approaches 0. As it approaches 0 from the left, it travels towards negative infinity. As it approaches 0 from the right, it travels towards positive infinity. Therefore, the limit of the function as it approaches 0 does not exist.
0
What is the limit as x approaches infinity of the square root of x? Ans: As x approaches infinity, root x approaches infinity - because rootx increases as x does.
When any number is divided by infinity, the result approaches zero but never actually reaches it. This is because infinity is not a specific number but rather a concept representing unboundedness. Mathematically, the limit of any finite number divided by infinity as infinity approaches infinity is zero.
The limit does not exist.
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.
1
When the limit as the function approaches from the left, doesn't equal the limit as the function approaches from the right. For example, let's look at the function 1/x as x approaches 0. As it approaches 0 from the left, it travels towards negative infinity. As it approaches 0 from the right, it travels towards positive infinity. Therefore, the limit of the function as it approaches 0 does not exist.
When we divide 1 by infinity, we are essentially taking the limit of 1 as the denominator approaches infinity. In mathematics, this limit is equal to zero. This is because as the denominator becomes infinitely large, the value of the fraction approaches zero. Therefore, 1 divided by infinity equals 0.
0
Infinity.
When the limit of x approaches 0 x approaches the value of x approaches infinity.
Limit as x tends to ∞: x/e^xAs you can see, as x approaches infinity, the sum becomes ∞/∞. Now we use l'Hospitals rules.d/dx(x) = 1 (Derivative)d/dx(e^x) = e^x (Derivative)therefore, the sum can be written as lim x tends to ∞ 1/e^xNow as x approaches infinity, the sum = 1/∞ = 0Therefore, lim x tends to infinity: x/e^x = 0
As X approaches infinity it approaches close as you like to 0. so, sin(-1/2)