How do you evaluate # x^(ln(7)/(1+ln(x)) # as x approaches infinity?

Question

Emilia Aguilar · Accepted Answer

7

Let #y=x^(ln 7 /(1+ln x))#. #ln y = ln 7/(1+ln x) ln x# #lim x to oo# of # ln y# # = ln 7 (lim x to oo# of #ln x/(1+ln x))# Apply L'Hospital rule. The limit of #ln y = ln 7 lim x to oo# of #(1/x)/(1/x)# #=ln 7 (1) So, #lim x to oo# of #y = 7#.

Samantha Adkison · Answer

undefined As x approaches infinity, the expression x^(ln(7)/(1+ln(x))) can be evaluated using the limit properties. By applying the limit rules, we can simplify the expression. Taking the natural logarithm of both sides, we get ln(y) = ln(7)/(1+ln(x)) * ln(x). Rearranging the equation, we have ln(y) * (1+ln(x)) = ln(7) * ln(x). Expanding the left side, we get ln(y) + ln(y) * ln(x) = ln(7) * ln(x). Dividing both sides by ln(x), we obtain ln(y)/ln(x) + ln(y) = ln(7). As x approaches infinity, ln(x) also approaches infinity. Therefore, ln(y)/ln(x) approaches 0. Thus, the equation simplifies to 0 + ln(y) = ln(7). Finally, solving for ln(y), we find ln(y) = ln(7). Taking the exponential of both sides, we get y = e^(ln(7)), which simplifies to y = 7. Therefore, as x approaches infinity, the expression x^(ln(7)/(1+ln(x))) approaches 7.