so im really stuck on this limit problem. I think it would be ne
lim (2x-1)/(lxl-3)
x-> infinity (and also negative infinity)
for x>0, (2x-1)/(|x|-3) = (2x-1)/(x-3) → 2
for x<0, (2x-1)/(|x|-3) = (2x-1)/(-x-3) → -2
Review the actual definition of |x|
Go to wolframalpha.com and type in your function to see the graph.
oops i accidentally submitted it. I meant to finish it by saying would it not exist for either of them?
ohhh ok. Wow I can't believe I didn't get that. Thank you so much!! :)
To solve the limit problem lim (2x-1)/(|x|-3) as x approaches both positive and negative infinity, we need to consider the behavior of the function as x gets larger and larger or smaller and smaller.
First, let's evaluate the limit as x approaches positive infinity:
As x approaches positive infinity, the absolute value |x| is equal to x since x is positive. So we can rewrite the expression as lim (2x-1)/(x-3).
To determine the limit as x approaches positive infinity, we can divide every term in the expression by the highest power of x, which is x in this case.
lim (2x-1)/(x-3) = lim (2 - 1/x)/(1 - 3/x) (dividing by x)
Next, we take the limit as x approaches positive infinity of the simplified expression:
lim (2 - 1/x)/(1 - 3/x) = (2 - 0)/(1 - 0) = 2/1 = 2
Therefore, the limit as x approaches positive infinity is 2.
Now, let's evaluate the limit as x approaches negative infinity:
As x approaches negative infinity, the absolute value |x| is equal to -x since x is negative. So we can rewrite the expression as lim (2x-1)/(-x-3).
Again, we divide every term in the expression by the highest power of x, which is -x in this case.
lim (2x-1)/(-x-3) = lim (2 + 1/x)/(1 + 3/x) (dividing by -x)
Next, we take the limit as x approaches negative infinity of the simplified expression:
lim (2 + 1/x)/(1 + 3/x) = (2 + 0)/(1 + 0) = 2/1 = 2
Therefore, the limit as x approaches negative infinity is also 2.
In summary, the limit of (2x-1)/(|x|-3) as x approaches both positive and negative infinity is 2.