Evaluate the following:

1. \displaylines{lim \\ x \to 0} \dfrac{sinax}{x}

Solution:
Given function takes indeterminate (0/0) form at x = 0. So,

\displaylines{lim \\  x \to 0} \dfrac{sinax}{x} * \dfrac{a}{a}

= a * \displaylines{lim \\ x \to 0} \dfrac{sinax}{ax}

[\because \displaylines{lim \\x \to 0} \dfrac{sinx}{x} = 1]

= a * 1

= a


2. \displaylines{lim \\ x \to 0} \dfrac{tan bx}{x}

Solution:
Given function takes indeterminate (0/0) form at x = 0. So,

\displaylines{lim \\ x \to 0} \dfrac{sinbx}{cosbx} * \dfrac{1}{x}

= \displaylines{lim \\x \to 0} \dfrac{sinbx}{cosbx} * \dfrac{b}{b} * \dfrac{1}{x}

= b * \displaylines{lim \\x \to 0} \dfrac{sinbx}{bx} * \dfrac{1}{cosbx}

[\because \displaylines{lim \\x \to 0} \dfrac{sinx}{x} = 1]

= b * 1 * \dfrac{1}{cos0}

= b * 1 * \dfrac{1}{1}

= b


3. \displaylines{lim \\ x \to 0} \dfrac{\sin mx}{\sin nx}

Solution:
Given function takes indeterminate (0/0) form at x = 0. So,

\displaylines{lim \\ x \to 0} \dfrac{\sin mx}{\sin nx} * \dfrac{mx}{mx} * \dfrac{nx}{nx}

= \displaylines{lim \\ x \to 0} \dfrac{\sin mx}{mx} * \dfrac{m}{n} * \dfrac{1}{\frac{\sin nx}{nx} }

= \dfrac{m}{n} * \displaylines{lim \\ x \to 0} \dfrac{\sin mx}{mx} * \dfrac{1}{\frac{\sin nx}{nx}}

[\because \displaylines{lim \\x \to 0} \dfrac{sinx}{x} = 1]

= \dfrac{m}{n} * 1 * \dfrac{1}{1}

= \dfrac{m}{n}


4. \displaylines{lim \\ x \to 0} \dfrac{tan ax}{tan bx}

Solution:
Given function takes indeterminate (0/0) form at x = 0. So,

\displaylines{lim \\ x \to 0} \dfrac{sinax}{cosax} * \dfrac{cos bx}{sin bx}

= \displaylines{lim \\x \to 0}\dfrac{sinax}{sinbx} * \dfrac{cosbx}{cosax}

= \displaylines{lim \\ x \to 0}\dfrac{sinax}{sinbx} * \displaylines{lim \\x \to 0}\dfrac{cosbx}{cosax}

= \displaylines{lim \\x \to 0} \dfrac{sinax}{ax} * \dfrac{1}{\frac{sin bx}{bx}} * \dfrac{ax}{bx} * \dfrac{cos0}{cos0}

= \displaylines{lim \\ x \to 0} \dfrac{sinax}{ax} * \dfrac{1}{\frac{sin bx}{bx}} * \dfrac{a}{b} * \dfrac{1}{1}

[\because \displaylines{lim \\x \to 0} \dfrac{sinx}{x} = 1]

= 1 * \dfrac{1}{1} * \dfrac{a}{b} * 1

= \dfrac{a}{b}


5. \displaylines{lim \\ x \to 0} \dfrac{\sin px}{\tan qx}

Solution:

Here,
The function takes indeterminate (0/0) form at x = 0.

So,
\displaylines{lim \\ x \to 0} (\sin px ) * \dfrac{ \cos qx}{\sin qx}

= \displaylines{lim \\x \to 0}\dfrac{\sin px{{px} * (\cos qx) * \dfrac{1}{\dfrac{\sin qx}{qx}} * \dfrac{px}{qx}

= \displaylines{lim \\x \to 0} \dfrac{\sin px}{px} * (\cos qx) * \dfrac{1}{\dfrac{\sin qx}{qx}} * \dfrac{p}{q}

[\because \displaylines{lim \\x \to 0} \dfrac{sinx}{x} = 1]

= 1 * cos 0 * \dfrac{1}{1} * \dfrac{p}{q}

= \dfrac{p}{q}


6. \displaylines{lim \\ x \to a} \dfrac{sin ( x-a)}{x^2 - a^2}

Solution:

Given,

\displaylines{lim \\ x \to a} \dfrac{sin ( x-a)}{x^2 - a^2}

= \displaylines{lim \\ x - a \to a - a} \dfrac{sin(x-a}{(x-a)} × \displaylines{lim \\ x \to a} \dfrac{1}{(x+a)}

= \displaylines{lim \\ x- a \to 0} \dfrac{sin(x-a)}{(x-a)} × \dfrac{1}{(a+a)}

= 1 × \dfrac{1}{2a}

= \dfrac{1}{2a}



7. \displaylines{lim \\x \to p} \dfrac{x^2 - p^2}{tan (x - p)}

Solution:

\displaylines{lim \\x \to p} \dfrac{x^2 - p^2}{tan (x - p)}

= \displaylines{lim \\ x \to p} \dfrac{(x+p)(x-p)}{tan(x-p)}

= \displaylines{lim \\ x - p \to p -p} \dfrac{(x-p)}{tan (x -p)} × \displaylines{lim \\ x \to p} (x +p)

= \displaylines{lim \\ x - p \to 0} \dfrac{1}{ \frac{tan(x-p)}{(x-p)}} × (p+p)

= \dfrac{1}{1} × (2p)

= 2p


8. \displaylines{lim \\x \to 0} \dfrac{\sin ax * \cos bx}{\sin cx}

Solution:

\displaylines{lim \\ x \to 0} \dfrac{\sin ax * \cos bx}{\sin cx}

= \displaylines{lim \\ x \to 0} \dfrac{\sin \frac{ax}{ax} × ax × \cos bx}{\sin \frac{cx}{cx} × cx}

= 1 × ax × cos(b×0) × \dfrac{1}{1 × cx}

= ax × 1 × \dfrac{1}{cx}

= \dfrac{a}{c}


9. \displaylines{lim \\x \to 0}\dfrac{1 - cosx}{x^2}

Solution:

Given,

\displaylines{lim \\x \to 0}\dfrac{1 - cosx}{x^2}

= \displaylines{lim \\ x \to 0} \dfrac{2sin^2 \frac{x}{2}}{x^2}

= 2× \displaylines{lim \\ x \to 0} \dfrac{sin^2 \frac{x}{2}}{ \left ( \frac{x}{2} \right)^2} × \dfrac{1}{2^2}

= 2× 1 × \dfrac{1}{2^2}

= \dfrac{1}{2}


10. \displaylines{lim \\x \to 0}\dfrac{1- \cos 6x}{x^2}

Solution:

Given,

\displaylines{lim \\x \to 0}\dfrac{1- \cos 6x}{x^2}

= \displaylines{ lim \\x \to 0} \dfrac{2 sin^2 3x}{x^2}

= 2× \displaylines{lim \\x \to 0} \dfrac{sin^2 3x}{(3x)^2} × 3^2

= 2 × 1 × 9

= 18


29. Find the limits of:

a) \displaylines{lim \\ x \to 0} \dfrac{e^{6x} - 1}{x}

Solution:

Given function takes (0/0) indeterminate form at x = 0 so,

\displaylines{lim \\ x \to 0} \dfrac{e^{6x} - 1}{x}

= \displaylines{lim \\ x \to 0} \dfrac{e^{6x} - 1}{6x} * 6

$= 6 * \displaylines{lim \\ x \to 0} \dfrac{e^{6x} - 1}{6x}$

= 6 * 1

[ \because \displaylines{lim \\ x \to 0} \dfrac{e^x -1}{x} = 1]

= 6


b) \displaylines{lim \\ x \to 0} \dfrac{e^{2x} - 1}{x . 2^{x+1}}

Solution:

Given function takes (0/0) indeterminate form at x = 0

\displaylines{lim \\ x \to 0} \dfrac{e^{2x} - 1}{x . 2^{x+1}}

= \displaylines{ lim \\ x \to 0} \dfrac{e^{2x} -1 }{x * 2^x * 2^1}

$= \displaylines{lim \\ x \to 0} \ left [ \dfrac{e^{2x} - 1}{2x } * \dfrac{1}{2^x} \right ]$

[ \because \displaylines{lim \\ x \to 0} \dfrac{e^x -1}{x} = 1]

= 1 * \dfrac{1}{2^0}

= 1 * 1

= 1


c) \displaylines{ lim \\ x \to 0} \dfrac{e^{ax} - e^{bx}}{x}

Solution:

Given function takes (0/0) indeterminate form at x = 0

\displaylines{ lim \\ x \to 0} \dfrac{e^{ax} - e^b{x} + 1 - 1}{x}

= \displaylines{ lim \\ x \to 0} \dfrac{(e^{ax} - 1) - (e^{bx} -1)}{x}

= \displaylines{ lim \\ x \to 0} \dfrac{e^{ax}-1}{ax}*a - \displaylines{ lim \\ x \to 0} \dfrac{e^{bx} -1}{bx} * b 

= (1 * a) - (1 * b)

= a - b