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Grade: 12th pass
        
Please help me with the question given in the image attachment   
one month ago

Answers : (1)

Samyak Jain
284 Points
							
Put x \approx 0 in the limit and you will find that it is of the form 1\infty.
lim   {f(x)}g(x)  =  e^(x\rightarrowalim{f(x) – 1}.g(x))
x\rightarrowa
So, lim [tan(\pi/4 + x)]1/x =e^ lim [tan(\pi/4 + x) – 1].1/x = e^ lim [{(tan(\pi/4) + tanx) / (1 – tan(\pi/4)tanx)} – 1] / x
     x\rightarrow0                                                                       x\rightarrow0
= e^ lim (1 + tanx – 1 + tanx)/x(1 – tanx) = e^ lim 2tanx / x(1 – tanx) = e^ 2 lim (tanx / x) / lim (1 – tanx)
      x\rightarrow0                                                x\rightarrow0                                    x\rightarrow0               x\rightarrow0
 = e^ 2.1/(1 – 0)  = e2 .
Here use lim tanx / x  =  1  and  tan(A+B) = (tanA + tanB) / (1 – tanAtanB)
             x\rightarrow0
one month ago
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