GATE (TF) Textile 2010 Question Paper Solution | GATE/2010/TF/46

When $x\rightarrow 4$ , $\lim \frac{x^3-64}{log_e(x-3)}$ will be equal to

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$x\rightarrow 4$ , $\lim \frac{x^3-64}{log_e(x-3)}$

We will expand the $log(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}-.........$

$x\rightarrow 4$ , $\lim \frac{(x-4)(x^2+16+4x)}{(x-4)-\frac{(x-4)^2}{2}+\frac{(x-4)^3}{3}+......}$

$x\rightarrow 4$ , $\lim \frac{(x^2+16+4x)}{1-\frac{(x-4)}{2}+\frac{(x-4)^2}{3}+......}$

By taking limit-

$\frac{(4^2+16+4\times 4)}{1-\frac{(4-4)}{2}+\frac{(4-4)^2}{3}+......}$

16+16+16=48