Solve the limit

Question

lim(x--->0^+) (e^x+x)^2/x

Answer 1

The limit has the indeterminate form 1^∞

Let y = (e^x+x)^2/x

Then lny = (2/x)ln(e^x+x) = 2ln(e^x+x) / x. Taking the limit as x approaches 0 from the right, we get the indeterminate form 0/0.

So, by L'Hopital's Rule, the limit is equivalent to the limit as x approaches 0 from the right of 2(e^x+1) / (e^x+x)

= 2(2) / 1 = 4

Therefore, we have lim_x→0⁺ lny = 4

So, the given limit = lim_x→0⁺y = lim_x→0⁺(e^lny) = e⁴

^

Answer 2

if possible, try to graph the expression f(x) = y = (e^x +x)^2/x

it appears to be similar in shape to a rectanglar hyperbola with two separate branches in quadrants I and III

sort of two separate parabola like shapes. the quadrant I branch upward opening, the quadrant III branch downward opening

use an online graphing calculator, such as Desmos

graphically, it appears to be discontinuous at x=0 with the y axis as the asymptote

with the limit as x approaches 0+ as + infinity

and the limit as x approaches 0- as - infinity

if you plug in x=0 into the expression you get division by zero which is undefined, so you would often use L'Hopital's rule and take the derivative of numerator and denominator separately to get

2(e^x +x)(e^x+1)/1 = 2(e^x+x)(e^x+1)

use the chain rule let u=e^x +x, then f'(u) = 2uu' with u'=e^x +1

then plug in x=0 to get

2(1+0)(1+1) = 2(1)(2) = 4

limit as x approaches 0 of (e^x+x)^2/x would then seem to be = 4

but that can't be correct. There are cases where L'Hopital's rule cannot be applied. see the wikipedia article.

this example is one of those cases.

or

try some of the online L'Hopital calculators

and derivative calculators

a graphing calculator solves the problem