Evaluate the limit #lim_(x-0) (e^x+3x)^(1/x) #?

Question

Evaluate the limit #lim_(x->0) (e^x+3x)^(1/x) #?

Cameron Alexander · Accepted Answer

#e^4# As we know #e^x = 1+x+O(x^2)# then #lim_(x->0)(e^x+3x)^(1/x) = lim_(x->0)(1+x+3x+O(x^2))^(1/x) = lim_(x->0)(1+x+3x)^(1/x)# now making #y = 4x# #lim_(x->0) (e^x+3x)^(1/x) =lim_(y->0)(1+y)^(4/y) = (lim_(y->0)(1+y)^(1/y))^4 = e^4#

Emma Aldridge · Answer

#lim_(x->0) (e^x+3x)^(1/x) = e^4#

Write the function as:

#(e^x+3x)^(1/x) = (e^(ln(e^x+3x)))^(1/x) = e^(ln(e^x+3x)/x)#

Consider now the limit:

#lim_(x->0) ln(e^x+3x)/x#

It is in the indeterminate form #0/0# so we can use l'Hospital's rule:

#lim_(x->0) ln(e^x+3x)/x = lim_(x->0) (d/dx ln(e^x+3x))/(d/dx x)#

#lim_(x->0) ln(e^x+3x)/x = lim_(x->0) (e^x+3)/(e^x+3x) = 4#

As the limit is finite and the function #e^x# is continuous for #x in RR# we have:

#lim_(x->0) e^(ln(e^x+3x)/x) = e^((lim_(x->0) ln(e^x+3x)/x)) = e^4#

Christopher Agresta · Answer

undefined The limit of (e^x+3x)^(1/x) as x approaches 0 is e^3.

Elijah Abram · Answer

undefined To evaluate the limit lim_(x->0) (e^x+3x)^(1/x), we can rewrite the expression using properties of logarithms and exponential functions:

lim_(x->0) (e^x+3x)^(1/x) 
= lim_(x->0) e^(1/x * ln(e^x + 3x))

Now, we can use the fact that ln(a^b) = b * ln(a):

= lim_(x->0) e^(ln(e^x + 3x)/x)

Applying L'Hôpital's Rule, where we differentiate the numerator and denominator separately:

= lim_(x->0) (d/dx(ln(e^x + 3x)))/(d/dx(x))

= lim_(x->0) ((e^x + 3)/(e^x + 3x))/(1)

Now, substitute x = 0 into the expression:

= ((e^0 + 3)/(e^0 + 3*0))/(1)
= (1 + 3)/(1)
= 4

So, the limit lim_(x->0) (e^x+3x)^(1/x) is equal to 4.

Evaluate the limit #lim_(x-&gt;0) (e^x+3x)^(1/x) #?

Evaluate the limit #lim_(x->0) (e^x+3x)^(1/x) #?