evaluate the limit
lim x ➟ ꝏ
(5x^2+3x)/(7x^4+2x^2+10)^1/2
how would I got about solving this? you can factor out an x on top but the bottom cant be factored (i dont think). the highest power in the denominator is x^4 but when you divided the top and the bottom by it then you still dont get a proper answer.
lim x ➟ ꝏ
(5x^2+3x)/(7x^4+2x^2+10)^1/2
the lead coefficient of (7x^4+2x^2+10)^1/2 will be √7 x^2
so lim x ➟ ꝏ
(5x^2+3x)/(7x^4+2x^2+10)^1/2 will converge to 5/√7 or (5√7) / 7
Well, when it comes to limits, sometimes it's all about finding clever ways to simplify the expression. Let's give it a shot!
First, we can observe that both the numerator and the denominator have a highest power of x². So, let's divide the entire expression by x²:
lim x ➟ ꝏ (5x^2+3x)/(7x^4+2x^2+10)^1/2 = lim x ➟ ꝏ (5+3/x)/(7+2/x^2+10/x^4)^1/2
Now, as x approaches infinity (or the clown limbo, ꝏ), those pesky 3/x, 2/x², and 10/x⁴ terms all go to zero. We can let them go on a vacation to the land of "negligible numbers"!
This leaves us with:
lim x ➟ ꝏ (5+0)/(7+0+0)^1/2
Simplifying further:
lim x ➟ ꝏ (5)/(7)^1/2
And finally:
lim x ➟ ꝏ (5)/(√7)
So, the limit of the expression as x goes to infinity is 5 divided by the square root of 7, which is approximately 1.88.
But remember, my dear friend, math can sometimes be as unpredictable as a banana peel on the floor. So, proceed with a dash of caution!
To evaluate the given limit, lim x ➟ 0 [(5x^2+3x)/(7x^4+2x^2+10)^1/2], you are correct that you cannot directly factor the denominator.
However, you can simplify the expression using algebraic manipulations. Let's go through the steps:
Step 1: Start by factoring out the highest power of x from both the numerator and denominator. In this case, the highest power of x is x^2. So, we can rewrite the expression as:
lim x ➟ 0 [x^2 * (5 + 3/x) / ((7x^4 + 2x^2 + 10)^1/2)]
Step 2: Next, simplify the denominator. Notice that we have (7x^4 + 2x^2 + 10)^1/2. To separate the square root from the expression, we can rewrite it as (7x^4 + 2x^2 + 10)^1/2 = [(7x^4 + 2x^2 + 10) / x^4]^1/2. Now the denominator becomes [(7x^4 + 2x^2 + 10) / x^4]^1/2.
Step 3: Simplify the expression within the square root:
(7x^4 + 2x^2 + 10) / x^4
Step 4: Take the square root of the expression within the square root:
[(7x^4 + 2x^2 + 10) / x^4]^1/2 = [(7x^4 + 2x^2 + 10)^1/2 / x^2]
Step 5: Substitute the simplified expressions back into the original limit:
lim x ➟ 0 [x^2 * (5 + 3/x) / [(7x^4 + 2x^2 + 10)^1/2 / x^2]]
Step 6: Simplify further:
= lim x ➟ 0 [x * (5 + 3/x) / (7x^4 + 2x^2 + 10)^1/2]
Step 7: Now, substitute x = 0 into the expression:
= 0 * (5 + 3/0) / (7(0)^4 + 2(0)^2 + 10)^1/2
Step 8: Since we have an undefined term (3/0), the limit is undefined or does not exist.
Therefore, the limit lim x ➟ 0 [(5x^2+3x)/(7x^4+2x^2+10)^1/2] is undefined.
To evaluate the given limit, you can try applying some algebraic manipulation and the properties of limits. Here's how you can go about it:
1. Start by factoring out x from the numerator:
(5x^2 + 3x) / (7x^4 + 2x^2 + 10)^1/2
= x(5x + 3) / (7x^4 + 2x^2 + 10)^1/2
2. Next, observe that as x approaches infinity (∞) or negative infinity (-∞), the highest power in the denominator dominates the expression. In this case, the highest power in the denominator is x^4.
3. Divide the numerator and denominator by x^2 to cancel out the x^2 term in the denominator:
= x(5x + 3) / x^2(7x^2 + 2 + 10/x^2)^1/2
= (5x + 3) / x(7 + 2/x^2 + 10/x^4)^1/2
4. Now, as x approaches ∞ or -∞, the terms involving 1/x^2 and 1/x^4 tend to zero. Thus, you can simplify the expression further:
= (5x + 3) / x(7)^1/2
= (5 + 3/x) / (7)^1/2
5. Finally, you can evaluate the limit as x approaches ∞ or -∞. Since the denominator is a constant (7) and the numerator tends to 5, the overall result is:
= 5 / (7)^1/2
= 5 / √7
Therefore, the limit as x approaches ∞ or -∞ of (5x^2 + 3x) / (7x^4 + 2x^2 + 10)^1/2 is 5 / √7.
It's important to note that this method of evaluating limits is not the only approach, and depending on the given function, other techniques like L'Hôpital's Rule or the squeeze theorem may come into play.