Results 1 to 3 of 3

Math Help - find k in form of limit

  1. #1
    Member
    Joined
    Jun 2008
    Posts
    175

    Question find k in form of limit

    Question asks to find k in the form of a limit of
    <br />
\begin{array}{l}<br />
 y = 4^x  \\<br />
 \frac{{dy}}{{dx}} = k.4^x  \\<br />
 \end{array}<br />

    This is my attempt

    <br />
\begin{array}{l}<br />
 \frac{{dy}}{{dx}} = \mathop {\lim }\limits_{h \to 0} \frac{{4^{a + h}  + 4^a }}{h} \\<br />
 k.4^a  = \frac{{4^{a + h}  + 4^a }}{h} \\<br />
 k = \frac{1}{{4^a }}\frac{{4^{a + h}  + 4^a }}{h} \\<br />
 \end{array}<br />

    The answer is given as

    <br />
\mathop {\lim }\limits_{h \to 0} \frac{{4^h  + 1}}{h}<br />

    I'm not sure how to get to this.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hello,
    Quote Originally Posted by Craka View Post
    Question asks to find k in the form of a limit of
    <br />
\begin{array}{l}<br />
 y = 4^x  \\<br />
 \frac{{dy}}{{dx}} = k.4^x  \\<br />
 \end{array}<br />

    This is my attempt

    <br />
\begin{array}{l}<br />
 \frac{{dy}}{{dx}} = \mathop {\lim }\limits_{h \to 0} \frac{{4^{a + h}  {\color{red}-} 4^a }}{h} \\<br />
 k.4^a  = \frac{{4^{a + h}  {\color{red}-} 4^a }}{h} \\<br />
 k = \frac{1}{{4^a }}\frac{{4^{a + h}  {\color{red}-} 4^a }}{h} \\<br />
 \end{array}<br />

    The answer is given as

    <br />
\mathop {\lim }\limits_{h \to 0} \frac{{4^h  {\color{red}-} 1}}{h}<br />

    I'm not sure how to get to this.
    I guess these are - signs and not + (see the red parts). And you are very near to the solution !

    Note that 4^{a+h}-4^a=4^a \cdot 4^h-4^a=4^a \cdot \left(4^h-1\right)

    Can you see a simplification ?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jun 2008
    Posts
    175
    Yes sorry they should be negative signs

    <br />
\begin{array}{l}<br />
 \frac{{dy}}{{dx}} = \mathop {\lim }\limits_{h \to 0} \frac{{4^{a + h}  - 4^a }}{h} \\ <br />
 k.4^a  = \frac{{4^{a + h}  - 4^a }}{h} \\ <br />
 k = \frac{1}{{4^a }}\frac{{4^{a + h}  - 4^a }}{h} \\ <br />
 \end{array}<br />
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Limit of an indeterminant form 0/0
    Posted in the Calculus Forum
    Replies: 4
    Last Post: April 22nd 2010, 10:50 PM
  2. Limit of an Indeterminate Form Problem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 25th 2010, 05:58 PM
  3. Limit of an Indeterminate Form
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 23rd 2010, 04:41 PM
  4. Indeterminant limit form
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 7th 2009, 03:49 AM
  5. Limit with Indeterminate Form
    Posted in the Calculus Forum
    Replies: 2
    Last Post: February 22nd 2009, 12:27 PM

Search Tags


/mathhelpforum @mathhelpforum