how is the limit of x^2/(1-cos^2(x)) as x goes to 0 is 1. I know how to show this using l'hopital rule. But can you show this using the rule that limit of (1-cosx)/x=1 as x goes to 0 and limit of sinx/x=1 as x goes to 0
1 -cos^2 x=sin^2x
lim (x/sinx)(x/sinx)=lim x/sinx lim x/sinx=1
remember if lim a/b = 1, then lim b/a has to equal 1.
To evaluate the limit of the expression (x^2)/(1 - cos^2(x)) as x approaches 0 using the given rules, we can perform the following steps:
Step 1: Rewrite the denominator 1 - cos^2(x) as sin^2(x).
Since sin^2(x) can be expressed as (1 - cos^2(x)), we can substitute it into the expression.
(x^2)/(1 - cos^2(x)) = (x^2)/(sin^2(x))
Step 2: Rearrange the expression using the limit rules you mentioned.
(x^2)/(sin^2(x)) = (x^2)/(x^2 * (sin^2(x)/x^2))
Notice that the numerator and the denominator both have x^2. By canceling out the x^2 terms, the expression becomes:
(x^2)/(x^2 * (sin^2(x)/x^2)) = 1/(sin^2(x)/x^2)
Step 3: Apply the given limit rules.
We know that the limit of (sin(x))/x as x approaches 0 is 1:
lim(x->0) (sin(x))/x = 1
Therefore, we can also conclude that:
lim(x->0) (sin^2(x))/x^2 = 1
Step 4: Use the result obtained from Step 3 to evaluate the original limit.
lim(x->0) (x^2)/(1 - cos^2(x)) = lim(x->0) 1/(sin^2(x)/x^2)
= 1/1 (since sin^2(x)/x^2 approaches 1 as x approaches 0)
= 1
Hence, the limit of the expression (x^2)/(1 - cos^2(x)) as x approaches 0 is indeed 1.