I am studying for a final exam and in our review packet we have this question, which I am having trouble with:
lim (x-->0) ((e^x)+x)^(1/x)
I tried this problem two ways.
The first was with L'Hopitals rule...
lim(x-->0) (1/x)((e^x)+x)^(1/x-1) * (e^x+1)
but that doesn't seem to help because I don't know what to do with (1/x-1)
Then I tried it like this:
y= ((e^x)+x)^(1/x)
ln y= ln((e^x)+x)^(1/x)
ln y= (1/x)ln((e^x)+x)
and then apply l'hopitals rule (because the above yields 0/0... but I always end up with an x in the denominator.
How do I solve this?
keep going
ln y= 1/x * ln (e^x+x)
again= 1/1 * 1/(e^x+x)* (e^x+1)
and that limit is 2
so since you evaluated the limit of ln y,
then lim y must be e^2
To solve the limit of ((e^x)+x)^(1/x) as x approaches 0, we can take a different approach. Let's go step by step.
First, let's rewrite the expression using the natural logarithm function (ln):
y = ((e^x)+x)^(1/x)
ln(y) = ln(((e^x)+x)^(1/x))
Now, let's simplify the expression using properties of logarithms:
ln(y) = (1/x)ln((e^x)+x)
Next, let's differentiate both sides of the equation using the chain rule:
(d/dx) ln(y) = (d/dx) [(1/x)ln((e^x)+x)]
The derivative of ln(y) with respect to x can be written as:
(1/y)(dy/dx)
For the right-hand side of the equation, we need to apply the product rule and chain rule:
(d/dx) [(1/x)ln((e^x)+x)] = (1/x)(d/dx)[ln((e^x)+x)] + ln((e^x)+x)(d/dx)(1/x)
Now, let's simplify the right-hand side:
(1/x)(d/dx)[ln((e^x)+x)] + ln((e^x)+x)(d/dx)(1/x)
= (1/x)(d/dx)[ln((e^x)+x)] - ln((e^x)+x)/x^2
To evaluate the derivative (d/dx)[ln((e^x)+x)], we can use the chain rule:
(d/dx)[ln((e^x)+x)] = (1/((e^x)+x))((d/dx)[((e^x)+x)])
Now, let's simplify (d/dx)[((e^x)+x)]:
(d/dx)[((e^x)+x)] = (d/dx)[(e^x)] + (d/dx)[x]
= e^x + 1
Plugging this back into the equation:
(1/x)(d/dx)[ln((e^x)+x)] - ln((e^x)+x)/x^2
= (1/x)(1/((e^x)+x))(e^x + 1) - ln((e^x)+x)/x^2
= (e^x + 1)/((x^2)((e^x)+x)) - ln((e^x)+x)/x^2
Now, let's evaluate this expression as x approaches 0:
lim (x-->0) [(e^x + 1)/((x^2)((e^x)+x)) - ln((e^x)+x)/x^2]
= [(e^0 + 1)/((0^2)((e^0)+0)) - ln((e^0)+0)/0^2]
= (1/(0(1+1)) - ln(1)/0
= (1/0) - ln(1)/0
At this point, we have an indeterminate form (0/0). To simplify it further, we can use L'Hopital's rule.
Take the derivative of the numerator and denominator separately:
(d/dx) [(e^x + 1)/((x^2)((e^x)+x))] = (d/dx) (e^x + 1) / (x^2)(e^x + x) + ((e^x + 1)(d/dx)[(x^2)(e^x + x)])
= e^x / (x^2)(e^x + x) + (1/x^2) * (2x(e^x + x) + x^2(e^x + 1))
= e^x / (x^2)(e^x + x) + (2x(e^x + x) + x^2(e^x + 1)) / x^2
Now, let's evaluate this expression as x approaches 0:
lim (x-->0) [e^x / (x^2)(e^x + x) + (2x(e^x + x) + x^2(e^x + 1)) / x^2]
= [e^0 / (0^2)(e^0 + 0) + (2(0)(e^0 + 0) + 0^2(e^0 + 1)) / 0^2]
= [1 / (0)(1) + 0 / 0]
= Infinity
Therefore, the limit of ((e^x)+x)^(1/x) as x approaches 0 is Infinity.
To solve this limit, you can use a different approach. By analyzing the form of the limit, you can see that it is an indeterminate form of the type 1^∞. In such cases, it is helpful to take the natural logarithm of the expression and convert it into a different form. Here's how you can solve it step-by-step:
1. Start by taking the natural logarithm of the function:
ln(y) = ln[((e^x) + x)^(1/x)]
2. Next, use the properties of logarithms to simplify the expression:
ln(y) = (1/x) * ln((e^x) + x)
3. Now we need to evaluate the limit of ln(y). As x approaches 0, ln(y) will approach a certain value, which is the natural logarithm of the limit we're trying to find.
4. Applying the limit to the equation, we get:
lim (x->0) ln(y) = lim (x->0) [(1/x) * ln((e^x) + x)]
5. We still have an indeterminate form, so we can simplify further by applying L'Hôpital's rule. Differentiate both the numerator and denominator with respect to x:
lim (x->0) [(1/x) * ln((e^x) + x)] = lim (x->0) [ln((e^x) + x) - (1/x^2) * (e^x)/(e^x + x)]
6. Now we can evaluate the limit as x approaches 0:
lim (x->0) ln((e^x) + x) - (1/x^2) * (e^x)/(e^x + x)
7. Substitute x = 0 into the expression:
ln(1) - (1/0^2) * (e^0)/(e^0 + 0)
8. Simplify further:
ln(1) - (1/0) * (1)/(1)
9. ln(1) is equal to 0, and any number divided by 0 is undefined. Therefore, the limit is undefined.
In conclusion, the limit as x approaches 0 of ((e^x) + x)^(1/x) is undefined.