Lim x->0 tan^-1x/x solve?

2 Answers
Jan 23, 2018

#1#

Explanation:

#lim_(x->0) tan^(-1)x/x->0/0#

So we can try L'Hopital's rule:

For the numerator:

#d/dx tan^-1x=1/(1+x^2)#

and the denominator:

#d/dx(x) = 1#

There fore we get:

#lim_(x->0) tan^(-1)x/x=lim_(x-> 0 )(1/(1+x^2))/1=(1/1)/1=1#

Jan 23, 2018

Alternate way of thinking of it, but #underline("less rigorous") #

#lim_(x to 0) ( tan^(-1)(x) ) /x = 1 #

Explanation:

One thing to consider is approximations of trigonometric functions when #x# is particulaly small

The first thing to note is that #color(red)(tanx approx x # for #x# being small

This is due to #cosx approx 1 # and #sinx = x # for #x# being small

Hence #tanx = sinx / cosx approx x / 1 approx x #

So hence #tan^(-1)x # is simply #tanx# reflected in the line #y = x #

But we know the inverse of #y = x # is just # y = x #

#=> tan^(-1)(x) approx x # for small #x#

#=> tan^(-1)x / x approx x/x approx 1# when #x# is very small

#lim_(x to o) (tan^(-1)x )/ x = lim_(x to 0 ) 1 #

# = 1 #

We can say this as #x# gets small, to be particular, #x to 0 # but we #color(blue)(underline ("couldn't")# do this if we had the limit of something like #x to 1/10 #

Like I said at the beggining this method is a very non- rigorous approach, and doesnt always work on all functions

I'd advise you to use #L'H # rule for this problem, as answered prior