Calculus Limits - 2 Questions

2) lim (sin²(5x)) / (4x²) as x-->0
5) lim (abs(x-5)) / (25-x²) as x-->5 from the left
(abs = absolute value)
can someone explain how to do these problems

2) L'Hôpital's rule or without it:
lim (sin²(5x) / 4x²) = lim [sin(5x) / 2x]²
(5/2)² lim [sin(5x) / 2(5/2)x]²
(5/2)² lim [sin(5x) / 5x]²
(5/2)² [lim (sin(5x) / 5x)]²
(5/2)² (1)²
25/4

The fact that sin(ax) / (ax) -> 1 as x -> 0 is well known and proven by the squeeze theorem.


5) Use the definition of absolute value to observe that because x < 5, x - 5 < 0 and so
lim (5 - x) / (25 - x)
lim (5 - x) / (5 - x)(5 + x)
lim 1 / (5 + x)
1 / 10

2) lim (sin²(5x)) / (4x²) as x-->0
At a minimum you need to know the limit of {sinx/x} or its reciprocal as x-->0 is 1.

Your lim (sin²(5x)) / (4x²) can be broken up thus: lim (1/4) [sin 5x/x] [sin 5x/x] and
then the factors in brackets [ ] can be made to look like that top expression in braces { }
if we multiply top and bottom of the fraction by 5²; after which, taking the limit is rather
easy (5²/4). [Having to type fractions in a single line makes it difficult to display the simplicity
of this approach because so much additional effort is required to see all the numerators
and denominators readily.]

|a-b| =|b-a|
makes 5) lim (abs(x-5)) / (25-x²) as x-->5 from the left be
lim |5-x|/ [(5-x)(5+x)] as x-->5⁻

The fraction's numerator will be positive (because of the abs value sign)
the factor (5-x) will be positive too as long as x-->5⁻ (from smaller than 5)
thus those 2 factors will be +1 as long as x ≠ 5. That leaves a fraction-factor
[1/(5+x)] whose limit is easy to determine: 1/10

2) lim (sin^2 (5x))/(4x^2) = lim [(sin 5x)/(2x)]^2 --> [0/0]
Use l'Hospital's Rule.
lm (5 cos 5x/2)^2 = (5/2)^2 = 25/4

5) lim |x - 5|/(25 - x^2) = lim |x - 5|/[(5 - x)(5 + x)] = - lim (5 + x) --> -10