12) \lim_{x\rightarrow 0} \frac{a^{tanx}-a^{sinx}}{tanx-sinx}
3) \lim_{n\rightarrow \propto } \left[\frac{2}{\Pi } (n+1)\cos^{-1} \left(\frac{1}{n} \right)-n \right]
\lim_{x\rightarrow \propto } (x- ln coshx) where cosh t=\frac{e^{t}+e^{-t}}{2}
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15 Answers
1
106for second one, take asinx common from numerator and ull get of the form k*ax-1x
14) \lim_{x\rightarrow 0} \frac{tan^{2}x - x^{2}}{x^{2}tan^{2}x}
btw how to put space in latex?
11 > In the solution below , the limit tends to infinity in each step .
x - ln cosh x
= ln e x - ln cosh x
= ln e xcosh x
= ln 2 e xe x + e - x
= ln 21 + e - 2 x . . . . . . . . . . . . After Dividing By " e x " ;
Now , " ln " is a continuos function , hence we can take the limit inside the " ln " .
By taking the limit inside , i . e , by making x tending to infinity , the fraction inside " ln " tends to 2 as then " e - 2 x " tends to zero .
Hence , the answer is " ln 2 = . 6 9 3 " .
34) Lt {[(tanx-x) (tanx+x)]/x4} [x/tanx]2
x→0
now apply the expansion of tanx and then get the answer
i hope u can proceed now
62sECOND Question
\frac{a^{tan x}-a^{sin x}}{tan x- sin x}=a^{sin x}\frac{a^{tan x - sin x}-1}{tan x- sin x}=1.ln a=ln a
6use l'hospital for 2nd one....is it not coming by l'hospital?