Evaluate the following limits after having identified it's indeterminate form:
lim x->0+ (xe^(2x) +1)^(5/x)
My last one and I have no idea how to go about it. I was thinking L'H, but the derivative is long.
Thank you
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You will see solution step-by-step.
lim x->0+ (xe^(2x) +1)^(5/x) = 1^∞
so, take logs
log(lim) = lim(log) = lim (5/x) log(xe^(2x)+1)
Now we have ∞*0, so if we divide, that becomes
lim log(xe^(2x)+1) / (x/5)
Now L'H yields
= lim 5/(xe^(2x)+1) * e^(2x) (2x+1) = 5
so, now we have lim(log) = 5, so the original limit is e^5
To evaluate the limit, we first need to identify the indeterminate form. In this case, we have an expression of the form 0^∞, which is an indeterminate form.
Now, let's analyze how to solve this limit using a method other than L'Hôpital's Rule.
We can rewrite the expression using exponential properties:
lim x->0+ (xe^(2x) + 1)^(5/x)
= lim x->0+ [(xe^(2x))^(5/x)] * (1^(5/x))
Since 1 raised to any power is always 1, we can simplify the expression further:
= lim x->0+ [(xe^(2x))^(5/x)] * 1
= lim x->0+ (xe^(2x))^(5/x)
Next, we take the natural logarithm (ln) of both sides of the expression:
ln[lim x->0+ (xe^(2x))^(5/x)]
= ln[lim x->0+ x^(5/x) * e^(10/x)]
Now, use the logarithm properties. The logarithm of a product is the sum of the logarithms:
ln[lim x->0+ x^(5/x)] + ln[lim x->0+ e^(10/x)]
Now, let's evaluate each limit separately:
The first limit can still be rewritten as an indeterminate form, so we can apply L'Hôpital's Rule to solve it.
lim x->0+ x^(5/x) = lim x->0+ (e^(ln(x)))^(5/x)
= lim x->0+ e^[(5/x) * ln(x)]
Now take the derivative of the exponent separately:
[ d/dx (5/x) * ln(x) ]
= (5/x) * (d/dx ln(x)) + ln(x) * (d/dx (5/x))
= (5/x) * (1/x) + ln(x) * (-5/x^2)
= 5/x^2 - 5ln(x)/x^2
Now, evaluate the limit of the derivative:
lim x->0+ 5/x^2 - 5ln(x)/x^2
Substitute x = 0, since the limit is approaching 0:
lim x->0+ 5/0^2 - 5ln(0)/0^2
This simplifies to:
lim x->0+ 5/0 - undefined
Since we have an undefined value, the first limit does not exist.
Now, let's move on to the second limit:
lim x->0+ e^(10/x)
Since the argument of the exponential function goes to infinity, the limit evaluates to:
e^∞ = ∞
As a result, the second limit is infinity.
Now, combining both limits:
ln[lim x->0+ (xe^(2x))^(5/x)]
= ln[(lim x->0+ x^(5/x)) * (lim x->0+ e^(10/x))]
= ln[(undefined) * (∞)]
= ln[undefined]
Since the natural logarithm of an undefined value is also undefined, we can conclude that the original limit also does not exist.