Solve - x -4limtan x-1x - -4

x→π #160; 4 #160; lim ( tan x 1 ) #160; ( x π #160; 4 #160; ) #160; = #160; x→π #160; 4 #160; lim ( tan x 1 ) ( 1+tan ...

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Byju's Answer

Standard XII

Mathematics

Substitution Method to Remove Indeterminate Form

Solve : x →...

Question

Solve : $lim x \to \frac{π}{4} \frac{(tan x - 1)}{(x - \frac{π}{4})}$

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Solution

$l i m x \to \frac{π}{4} \frac{(tan x - 1)}{(x - \frac{π}{4})} =$ $l i m x \to \frac{π}{4} \frac{(tan x - 1) (1 + tan x)}{(x - \frac{π}{4}) (1 + tan x)}$

$= l i m x - \frac{π}{4} \to 0 \frac{(tan x - tan \frac{π}{4}) (1 + tan x)}{(x - \frac{π}{4}) (1 + tan \frac{π}{4} . tan x)}$

$= l i m x - \frac{π}{4} \to 0 \frac{(tan (x - \frac{π}{4})) (1 + tan x)}{(x - \frac{π}{4})}$

$= l i m x - \frac{π}{4} \to 0 \frac{(tan (x - \frac{π}{4}))}{(x - \frac{π}{4})} . l i m x - \frac{π}{4} \to 0 (1 + tan x)$

$= 1 \times (1 + tan \frac{π}{4})$

$= 1 + 1 = 2$

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Solve : limx→π4(tanx−1)(x−π4)

limx→π4(tanx−1)(x−π4)= limx→π4(tanx−1)(1+tanx)(x−π4)(1+tanx) =limx−π4→0(tanx−tanπ4)(1+tanx)(x−π4)(1+tanπ4.tanx)=limx−π4→0(tan(x−π4))(1+tanx)(x−π4)=limx−π4→0(tan(x−π4))(x−π4).limx−π4→0(1+tanx)=1×(1+tanπ4)=1+1=2

Solve : $lim x \to \frac{π}{4} \frac{(tan x - 1)}{(x - \frac{π}{4})}$