Evaluating required limit using algebraic manipulation, L' Hopital's rule and Maclaurin series

Lim-203
lim ((tan (3x))/3x)^(1/(1 - cos x)) when x tends to 0
#limitsandcontinuity #algebraic #manipulation #lhopital #rule #cipher
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[@Unknown.69762]

This is the limit of form 1^infinity so we will write it in e^limit
(tan3x/3x-1) 1/1-cosx
(tan3x-3x/tan3x) 1/1-cosx
Now multiply and divide by x³
(tan3x-3x/27x³) (3x/tan3x) ( x²/1-cosx) ×18
1/3×18=e⁶

[@cipherunity]

Good