How do you find the limit of #x Tan(9/x)# as x approaches infinity using l'hospital's rule?

Question

Charles Anctil · Accepted Answer

See below

#lim_(xrarroo)xtan(9/x)# has indeterminate form #oo*0#. In order to use l'Hospital's rule we rewrite the expression: #xtan(9/x) = (tan(9/x))/(1/x)#. The limit is the rewritten expression has form #0/0#, so we can apply l"Hospital. Differentiating the numerator and denominator gets us: #lim_(xrarroo)xtan(9/x) = lim_(xrarroo)(sec^2(9/x)(-9/x^2))/(-1/x^2)# # = lim_(xrarroo)(sec^2(9/x)(9))# # = 9sec^2(0) = 9# We do not need l'Hospital #xtan(9/x) = 9* sin(9/x)/(9/x) * 1/cos(9/x)# #lim_(urarroo)sinu/u=1#, so we get #lim_(xrarroo)xtan(9/x) = lim_(xrarroo) 9* sin(9/x)/(9/x) * 1/cos(9/x)# # = (9)(1)(1) = 9#

Charles Anctil · Answer

undefined To find the limit of x Tan(9/x) as x approaches infinity using L'Hospital's rule, we can apply the rule to the expression.

First, we differentiate both the numerator and denominator with respect to x. The derivative of x is 1, and the derivative of Tan(9/x) can be found using the chain rule.

Applying the chain rule, we have: 
d/dx(Tan(9/x)) = sec^2(9/x) * (-9/x^2)

Now, we can rewrite the expression as: 
lim(x->∞) [1 / (sec^2(9/x) * (-9/x^2))]

Next, we simplify the expression by taking the reciprocal of the denominator and multiplying it by the numerator: 
lim(x->∞) [(-x^2) / (9 * sec^2(9/x))]

As x approaches infinity, the term 9/x approaches 0. Therefore, we can rewrite the expression as: 
lim(x->∞) [(-x^2) / (9 * sec^2(0))]

Since sec^2(0) is equal to 1, the expression simplifies further to: 
lim(x->∞) [(-x^2) / 9]

As x approaches infinity, the numerator (-x^2) also approaches infinity. Therefore, the limit of x Tan(9/x) as x approaches infinity using L'Hospital's rule is negative infinity.