Find the limits:
lim {[1/(2+x)]-(1/2)}/x as x->0
lim (e^-x)cos x as x->infinity
...First off, I'm having trouble getting rid of complex fractions. I tried multiplying the whole numerator by [(2+x)/2], but you'd still get 0/undefined answer (the answer in the back of my book is -1/4)
...As for the second one, my guess is zero, since cosine of infinity would eventually reach zero.
(i need more help with the first one rather than the second)
Help please?
Thanks much!
To find the limit of the first expression, let's simplify it step by step.
Start with the expression:
lim {[1/(2+x)]-(1/2)}/x as x->0
Step 1: Simplify the numerator
Combine the fractions in the numerator by finding a common denominator, which is (2+x)(2):
= [2/(2+x) - (2+x)/(2(2+x))]/x
= [2 - (2+x)] / (2(2+x)) / x
= [-x] / (2(2+x)) / x
Step 2: Simplify the expression further
Cancel out the common factor of x:
= -1 / (2(2+x))
Step 3: Take the limit as x approaches 0
Substitute x = 0 into the expression:
= -1 / (2(2+0))
= -1 / 4
Therefore, the limit of the expression as x approaches 0 is -1/4.
Now let's move on to the second expression.
The limit of (e^-x)cos(x) as x approaches infinity:
As x approaches infinity, the exponential term e^-x will tend to 0 because the exponent becomes infinitely negative. Therefore, the expression simplifies to:
lim (e^-x)cos(x) as x->infinity = 0 * cos(infinity) = 0
So, your guess was correct. The limit of the second expression as x approaches infinity is indeed 0.
I hope this explanation helps you understand how to solve these limits. Let me know if you have any further questions!