\[\underset{x\to \infty }{\mathop{\lim }}\,\frac{{{(2x+1)}^{40}}{{(4x-1)}^{5}}}{{{(2x+3)}^{45}}}=\] [IIT 1990]

\[\underset{x\to \infty }{\mathop{\lim }}\,\frac{{{(2x+1)}^{40}}{{(4x-

1.	\[\underset{x\to \infty }{\mathop{\lim }}\,\frac{{{(2x+1)}^{40}}{{(4x-1)}^{5}}}{{{(2x+3)}^{45}}}=\] [IIT 1990]
A.	16
B.	24
C.	32
D.	8
Answer» D. 8

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\[\underset{x\to \infty }{\mathop{\lim }}\,\frac{{{(2x+1)}^{40}}{{(4x-1)}^{5}}}{{{(2x+3)}^{45}}}=\] [IIT 1990]

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