Evaluate the limit. Be sure to show your reasoning. limit as x approaches 0 (sin7x / tan3x) ?

I know that the bottom turns to sin3x/cos3x and then i multiply the recipricol in which I get the limit as x approaches 0 (sin x7)(cos3x / sin 3x) but I don't know where to go from here :(

3 Answers
Mar 6, 2018

#7/3#

Explanation:

#sin(7x)/tan(3x)#
#=((7x)+(7x)^3/3!...)/((3x)+(1/3)(3x)^3)...#(Maclaurin series)
#=(7+O(x^2))/(3+O(x^2))# (canceling #x#'s throughout)
#to 7/3# as #x to 0#

Mar 6, 2018

#7/3#

Explanation:

use L'Hôpital's Rule (also known as hospital's rule in English)
https://www.google.co.in/search?q=lhopital+rule&ie=&oe=

what this rule basically means is, if we have a limit for the division of two functions which looks like #0/0# or #oo/oo# which are indeterminate forms, we can divide the derivatives of the two functions to get the answer of the limit

therefore,
#lim_(x->c) f(x)/g(x) = lim_(x->c) (f'(x))/(g'(x))# if f(x) and g(x) would both be 0 or infinity or negative infinity if evaluated at c

therefore, taking the derivative of both the functions, we get
#lim_(x->0) sin(7x)/tan(3x) =lim_(x->0) (d/dx sin(7x))/(d/dx tan(3x))#
=

which is equal to

#lim_(x->0) (7cos(7x)) /(3 sec^2 (3x))#

now since #x# goes to 0 ,#7x# and #3x# would also approach 0

= #(7cos(0)) /(3 sec^2 (0))#

and #cos(0)# and #sec^2(0)# are both 1

therefore, the limit is
#7/3#
-

Mar 6, 2018

Without using derivatives, see below.

Explanation:

As you have said:

#(sin7x)/(tan3x) = (sin7x)/((sin3x)/(cos3x))#

# = (sin7x)/1 1/(sin3x) (cos3x)/1#

Now we'll write this so we can use
#lim_(trarr0)sint/t=1# and #lim_(trarr0)1/sint=1#

# = (sin7x)/(7x) * (7x)/(3x) (3x)/(sin3x) (cos3x)/1#

# =7/3 (sin7x)/(7x) (3x)/(sin3x) (cos3x)/1#

Now take limit as #xrarr0# (so both #3x# and #7x# also go to #0#)

#lim_(xrarr00)(sin7x)/(tan3x) = lim_(xrarr0)(7/3 (sin7x)/(7x) (3x)/(sin3x) (cos3x)/1#

# = 7/3 (1) (1) 1/1 = 7/3#