Question #6637d

Question

2441 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-25T19:54:40

#lim_(xrarrpi/4)(e^tanx-e)/(x-pi/4)=2e#

I'm assuming you meant the limit:

#lim_(xrarrpi/4)(e^tanx-e)/(x-pi/4)#

Trying to evaluate at #x=pi/4# gives the indeterminate form #0/0#. Thus, we can solve this limit using L'Hopital's rule.

#=lim_(xrarrpi/4)(d/dx(e^tanx-e))/(d/dx(x-pi/4))#

#=lim_(xrarrpi/4)(e^tanx(sec^2x))/1#

We can now evaluate the limit:

#=e^tan(pi/4)sec^2(pi/4)#

#=e^1(sqrt2)^2#

#=2e#

Answer 2 · 2017-03-02T09:28:13

#2e.#

We will use the following Standard Limits to evaluate the reqd.

limit :

# (SL1) : lim_(h to 0) (a^h-1)/h=lna, a gt 0.#

# (SL2) : lim_(theta to 0) tan theta/theta=1.#

Let, #l=lim_(x to pi/4) (e^tan x -e)/(x-pi/4).#

#=lim {e(e^(tanx-1) -1)}/(x-pi/4),#

#=lim {(e(e^(tanx-1) -1))/(tanx-1)}{(tanx-1)/(x-pi/4)},#

#=e(l_1l_2),.........(star)# where,

#l_1=lim_(x to pi/4) (e^(tanx-1)-1)/(tanx-1).#

Subst. #h=tanx-1," so that, as "x to pi/4, h to 0.#

#:. l_1=lim_(h to 0) (e^h-1)/h,

#=lne........[because, (SL1)],#

#:. l_1=1......................(star1).#

For, #l_2," we take, "x=pi/4+theta," so; when," x to pi/4, theta to 0.#

#:. l_2=lim_(theta to 0)(tan(pi/4+theta)-1)/theta#

#=lim {(1+tantheta)/(1-tantheta)-1}/theta#

#=lim (2tantheta)/{theta(1-tantheta)}#

#=lim (tantheta/theta)(2/(1-tantheta))#

#=(1)(2/(1-0)), ..................[because, (SL2)]#

#:. l_2=2...................(star2).#

#"From "(star1),(star2) & (star), "the reqd. lim. "l" follows as,"#

#l=2e.#

Enjoy Maths.!