Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Hello guys i have a homework question that once again has me stumped. I have a less then friendly teacher who thinks all her students are going to be the next Isaac Newton and she barely explains anything. So here is my question.

from reading my book it appears that i have two options. formula 1 says $$ \lim_{x\to a }\frac{f(x)-f(a)}{x-a}$$ and the second says $$ \lim_{h\to 0 }\frac{f(a+h)-f(a)}{h}$$

so i tried to use formula 2 to solve this problem

FIND The equation of the tangent line to the curve at the given point.

$$ \sqrt{x},$$ $$ (1,1)$$

step - 1 $$ \frac{f(1+h)-f(1)}{h}$$ step -2 $$\frac{\sqrt{1+h} - 1}{h}$$ step -3 used the conjugate pair to come up with $$\frac{\sqrt{1+h} - 1}{h} * \frac{\sqrt{1+h} + 1}{\sqrt{1+h} + 1} $$ step -4 the result of the above i would end up with $$ \frac{1+h-1}{h\sqrt{1+h}}$$ step 5 $$\frac{1}{\sqrt{1+0}} = 1$$

however the answer is $ \frac{1}{2}$ i cant see what I'm doing wrong and when should i use the first formula and when do i use the second formula? any help in answering this question would be most appreciated. Thanks Miguel

share|improve this question
3  
+1 for showing your work. –  N. S. Feb 15 '13 at 4:12
    
thank you N.S. i appreciate your answer and vote. –  Miguel Feb 15 '13 at 4:16
1  
$\frac 12$ is not a reasonable answer to a question that says "find the equation". It is a useful step (finding the correct slope) toward that. But +1 for showing your work-note (more for others than you) that the answers are more responsive because of that. –  Ross Millikan Feb 15 '13 at 4:23

4 Answers 4

up vote 1 down vote accepted

In step 4 you left off a $+1$ in one factor in the denominator. It should be $\displaystyle\frac{1+h-1}{h(\sqrt{1+h}+1)}$. Then $$\lim_{h\to 0}\frac{1+h-1}{h(\sqrt{1+h}+1)}=\lim_{h\to 0}\frac{1}{\sqrt{1+h}+1}=\lim_{h\to 0}\frac{1}{\sqrt{1+0}+1}={1\over 2}.$$

share|improve this answer
    
ahhhhh !!!! i feel so stupid! thank you. Thats where i made my mistake! –  Miguel Feb 15 '13 at 4:11
    
Not stupid... this is just "learning". ;-) –  JohnD Feb 15 '13 at 4:13

You lost a one going from step 3 to step 4. More exactly:

$$ \frac{\sqrt{1+h} - 1}{h} \frac{\sqrt{1+h} + 1}{\sqrt{1+h} {\bf+ 1}}=\frac{1+h - 1}{h(\sqrt{1+h} {\bf+ 1})}$$

share|improve this answer

Check step 4. You should have an extra h term being added in the denominator.

share|improve this answer

The answer from N. S. showed a problem going from step 3 to step 4. Going from step 4 to step 5 (using your step 4) you should say $\frac {1+h-1}{h\sqrt{1+h}}=\frac h{h\sqrt{1+h}}=\frac 1{\sqrt{1+h}}$ The multiplication by the conjugate was the correct approach to get rid of the $\frac 00$. If you couple these together and take the limit $h \to 0$, you will find the correct slope of $\frac 12$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.