1. Prove that $lim_{x\rightarrow \infty} \frac{ae^x + be^{-x}}{ce^x+de^{-x}}=\frac{a}{c} \quad [2<e<3, c \neq 0 ] $

Solution:

$lim_{x\rightarrow \infty} \frac{ae^x + be^{-x}}{ce^x+d e^{-x}} $

$=lim_{x\rightarrow \infty}  \frac{a + be^{-2x}}{c+d e^{-2x}} $

$=\frac{a + be^{-\infty}}{c+d e^{\infty}}$

$=\frac{a}{c} $  Proved

2. Prove that $lim_{x\rightarrow  - \infty} \frac{ae^x + be^{-x}}{ce^x+de^{-x}}=\frac{b}{d} \quad [2<e<3, d \neq 0 ] $

Solution:

$lim_{x\rightarrow  - \infty} \frac{ae^x + be^{-x}}{ce^x+de^{-x}}$

$=lim_{x\rightarrow  - \infty} \frac{ae^{2x} + b}{ce^{2x}+d}$

$\frac{ae^{-\infty} + b}{ce^{-\infty}+d}$

$=\frac{b}{d}$ Proved

4. Prove that $lim_{x\rightarrow 1} \frac{4^x-4}{x-1}=8log_e 2 $

Solution:

$lim_{x\rightarrow 1} \frac{4^x-4}{x-1}$ 

$= 4lim_{x-1 \rightarrow 0 } \frac{4(4^{x-1}-1)}{x-1}$

$=4 lim_{x-1 \rightarrow 0 } \frac{(4^{x-1}-1)}{x-1} $

$=4 log_e 4$

$=8log_e 2$ Proved

5. Prove that $lim_{x\rightarrow 2} \frac{log(2x-3)}{2(x-2)}=1$

Solution;

$lim_{x\rightarrow 2} \frac{log(2x-3)}{2(x-2)}$

$=lim_{2x\rightarrow 4} \frac{log(2x-4+1)}{2x-4})$

$=lim_{2x-4\rightarrow 0} \frac{log(2x-4+1)}{2x-4)}$

$=1$ Proved

6. Show that $lim_{x\rightarrow 0} ( 1+2x)^{\frac{1}{x}}=e^2$

Solution:

$lim_{x\rightarrow 0} ( 1+2x)^{\frac{1}{x}}$

$=lim_{x\rightarrow 0} ( 1+2x)^{\frac{1 \cdot 2}{ 2x}} $

$= \left[ lim_{x\rightarrow 0} ( 1+2x)^{\frac{1}{ 2x}} \right] ^2$

$=e^2$   proved            [Since $lim_{x\rightarrow 0}(1+x)^{\frac{1}{x}}=e $]

7. Show that $lim_{x\rightarrow 0} (1-3x)^ \frac{3}{x}= e^{-9}$]

Solution:

$lim_{x\rightarrow 0} (1-3x)^ \frac{3}{x}$

$=lim_{x\rightarrow 0} (1-3x)^{-\frac{1}{3x} \cdot ( -9)}$

$=\left[ lim_{-3x\rightarrow 0} (1-3x)^{- \frac{1}{3x}} \right] ^{-9} $

$=e^{-9}$  Proved        [Since $lim_{x\rightarrow 0}(1+x)^{\frac{1}{x}}=e $]

8. Show that $lim_{x\rightarrow 1}x^{\frac{1}{1-x}}=e^{-1}$

Solution:

$lim_{x\rightarrow 1}x^{\frac{1}{1-x}}$

$=lim_{x\rightarrow 1}(1+x-1)^{\frac{1}{1-x} \cdot (_1)}$

$= \left[ lim_{x-1 \rightarrow 0}(1+x-1)^{\frac{1}{1-x}} \right] ^{-1}$

$=e^{-1}$  Proved        [Since $lim_{x\rightarrow 0}(1+x)^{\frac{1}{x}}=e $]

9. Show That $lim_{x\rightarrow 0}(1+x)^{\frac{x+2}{x}}=e^{8}$

Solution:

$lim_{x\rightarrow 0}(1+x)^{\frac{x+2}{x}}$

$=lim_{x\rightarrow 0}(1+x)^{\frac{1}{4x} \cdot (4x+8)} $

$= { \left[ lim_{4x \rightarrow 0}(1+x)^{\frac{1}{4x}} \right] } ^{ lim_{x\rightarrow 0}(4x+8)}   $

$=e^{8}$    Proved        [Since $lim_{x\rightarrow 0}(1+x)^{\frac{1}{x}}=e $]