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

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Answer 1 · 2016-05-02T13:39:44

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#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#