How do you find the limit of #(1+7/x)^(x/10)# as x approaches infinity?

Question

Elijah Abram · Accepted Answer

There are other method available, but I use #lim_(mrarroo)(1+1/m)^m = e#

#(1+7/x)^(x/10) = ((1+7/x)^x)^(1/10) = [(1+1/(x/7))^(x/7)]^(7/10)#

As #xrarroo# the quotient #x/7 rarroo# as well.

So, with #m=x/7# we see that the expression in the brackets goes to #e#. That is

#lim_(mrarroo)[(1+1/(x/7))^(x/7)] = e#

Therefore,

#lim_(xrarroo)(1+7/x)^(x/10) = lim_(xrarroo)[(1+1/(x/7))^(x/7)]^(7/10)#

# = [lim_(xrarroo)(1+1/(x/7))^(x/7)]^(7/10)#

# = e^(7/10)#

Charles Arocha · Answer

undefined To find the limit of (1+7/x)^(x/10) as x approaches infinity, we can use the property of exponential limits.

First, we rewrite the expression as e^(ln((1+7/x)^(x/10))).

Next, we take the natural logarithm of both sides to simplify the expression further.

Using the properties of logarithms, we can rewrite the expression as (x/10) * ln(1+7/x).

Now, we can evaluate the limit as x approaches infinity.

As x approaches infinity, 7/x approaches 0, and ln(1+7/x) approaches ln(1) which is 0.

Therefore, the limit becomes (x/10) * 0, which is equal to 0.

Hence, the limit of (1+7/x)^(x/10) as x approaches infinity is 0.