How do you find the limit of #(1-e^(-2x))/(sec(x))# as x approaches 0?

1 Answer
Antoine
Mar 8, 2016

#0#

Explanation:

Step 1:
#(1-e^(-2x))/sec(x) rarr (1-e^(-2x))cos(x)#

because #sec(x)=1/cos(x)#

Step 2:
the cosine function is known to have a range of #-1<=cos(x)<=1#

So we can write:
#-1(1-e^(-2x))<=(1-e^(-2x))cos(x)<=1(1-e^(-2x))#

Step 3:
#lim_(xrarr0)-(1-e^(-2x)) rarr lim_(xrarr0)-1+1/e^(2x)= -1+1/e^(0)=-1+1/1=0#

Step 4:
#lim_(xrarr0)(1-e^(-2x)) rarr lim_(xrarr0)1-1/e^(2x)= 1-1/e^(0)=1-1/1=0#

Hence, #lim_(xrarr0)(1-e^(-2x))cos(x)=color(blue)0#

Recall:
According to the color(black)("Sandwich theorem"), if #a(x)<=f(x)<=b(x)#
And #lim_(x rarr c)a(x)=K=lim_(x rarr c)b(x)#
this implies that #lim_(xrarrc)f(x)= K#

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