Solving the Limit as x Approaches 0 of 1 – cos(x)/x: Simplification and Application of L’Hôpital’s Rule

limit as x approaches 0: 1-cosx/x

To solve the limit as x approaches 0 of the given expression 1 – cos(x)/x, we can first simplify it

To solve the limit as x approaches 0 of the given expression 1 – cos(x)/x, we can first simplify it.

Using the trigonometric identity, cos(x) = 1 – 2sin^2(x/2), we can rewrite the expression as:

1 – cos(x)/x = 1 – (1 – 2sin^2(x/2))/x
= 1 – 1/x + 2sin^2(x/2)/x

Next, let’s focus on the term 2sin^2(x/2)/x.

Using the double-angle identity, sin(θ/2) = √((1 – cosθ)/2), we can rewrite it as:

2sin^2(x/2)/x = 2(1 – cosx)/2x^2
= (1 – cosx)/x^2

Now, we can substitute the simplified expression back into the original limit:

1 – cos(x)/x = 1 – 1/x + (1 – cosx)/x^2

To evaluate the limit as x approaches 0, we can substitute the value of x = 0 into the expression:

1 – 1/0 + (1 – cos0)/0^2

However, we cannot divide by zero, so the original expression is undefined at x = 0. In this case, we cannot directly substitute 0 into the expression to find the limit.

To find the limit of the expression as x approaches 0, we can use L’Hôpital’s rule.

We will differentiate the numerator and denominator separately and then take the limit again:

Taking the derivative of the numerator:
d/dx (1 – cosx) = sinx

Taking the derivative of the denominator:
d/dx (x^2) = 2x

Now, we can evaluate the limit again:

lim(x->0) [(1 – cosx)/x^2] = lim(x->0) [sinx/2x]

Now, we can directly substitute x = 0 into the expression.

At x = 0,

lim(x->0) [sinx/2x] = sin(0)/2(0) = 0/0

Again, we encounter an undefined form. We can apply L’Hôpital’s rule once more.

Taking the derivative of the numerator:
d/dx (sinx) = cosx

Taking the derivative of the denominator:
d/dx (2x) = 2

Evaluating the limit again:

lim(x->0) [sinx/2x] = lim(x->0) [cosx/2] = cos(0)/2 = 1/2

Hence, the limit as x approaches 0 of the expression 1 – cos(x)/x is 1/2.

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