Show, using the properties of limits, that if lim x-->5 f(x)=3 then lim x---->5 x^2-4/f(x)=7
limit of quotient is quotient of limits, so we have
lim(x->5) x^2-4 = 25-4 = 21
lim(x->5) f(x) = 3
21/3 = 7
To show that if lim x-->5 f(x)=3, then lim x-->5 (x^2-4)/f(x)=7, we can use the properties of limits and algebraic manipulations.
First, let's rewrite the expression (x^2-4)/f(x) as a combination of two functions:
(x^2-4)/f(x) = x^2/f(x) - 4/f(x)
Now, let's focus on each part separately:
1) lim x-->5 x^2/f(x):
Since lim x-->5 f(x)=3, we can use the property of limits to write this as:
lim x-->5 x^2/f(x) = (lim x-->5 x^2) / (lim x-->5 f(x))
The limit of x^2 as x approaches 5 is 5^2=25. And using the given information, the limit of f(x) as x approaches 5 is 3. Thus,
lim x-->5 x^2/f(x) = 25/3
2) lim x-->5 4/f(x):
Similarly, since lim x-->5 f(x)=3, we have:
lim x-->5 4/f(x) = 4/(lim x-->5 f(x))
Plugging in the known limit, we get:
lim x-->5 4/f(x) = 4/3
Now, let's combine both parts:
lim x-->5 (x^2-4)/f(x) = lim x-->5 x^2/f(x) - lim x-->5 4/f(x)
= 25/3 - 4/3
= 21/3
= 7
Therefore, using the properties of limits, we have shown that if lim x-->5 f(x)=3, then lim x-->5 (x^2-4)/f(x)=7.
To show that lim x-->5 (x^2-4)/f(x)=7 using the properties of limits, we can start by using the limit algebra property that states:
If lim x-->c g(x) = L and lim x-->c h(x) = M, where both L and M are real numbers, then:
lim x-->c [g(x)/h(x)] = L/M, provided that M is not zero.
In this case, we know that lim x-->5 f(x) = 3. So, we have:
lim x-->5 (x^2-4)/f(x)
Using the limit algebra property, we can rewrite this expression as:
= lim x-->5 (x^2-4) / lim x-->5 f(x)
Since lim x-->5 f(x) = 3, we can substitute this value into the expression:
= lim x-->5 (x^2-4) / 3
Now, we can simplify the numerator by factoring:
= lim x-->5 (x+2)(x-2) / 3
Next, we can evaluate the limit of the expression as x approaches 5:
= [(5+2)(5-2)] / 3
= (7)(3) / 3
= 21 / 3
= 7
Therefore, we've shown that lim x-->5 (x^2-4)/f(x) = 7 using the properties of limits.