# Thread: Quick help in Limits

1. ## Quick help in Limits

$\displaystyle lim (h goes to 0) 5^h-2^h/h$

$\displaystyle lim (ygoes to -3) y^2-y-12/y+3$

again, help is much appreciated.

2. Hello,
Originally Posted by cjmac87

$\displaystyle lim (h goes to 0) 5^h-2^h/h$

$\displaystyle lim (ygoes to -3) y^2-y-12/y+3$

again, help is much appreciated.
I guess it's this :

$\displaystyle \lim_{h \to 0} \frac{5^h-2^h}{h}$

$\displaystyle \lim_{y \to -3} \frac{y^2-y-12}{y+3}$

(click on the pictures to get the code)

For the second one, note that $\displaystyle y^2-y-12=(y+3)(y-4)$

For the first one, are you familiar with L'Hospital's rule ?

3. Originally Posted by cjmac87

$\displaystyle lim (h goes to 0) 5^h-2^h/h$
i'm sure there's an easier way, and someone will point it out soon because everyone hates what i am about to suggest.

try applying L'Hopital's rule

*runs and jumps out the window to avoid the angry mob of mathematicians*

$\displaystyle lim (ygoes to -3) y^2-y-12/y+3$

again, help is much appreciated.
Hint: note that this is the same as $\displaystyle \lim_{y \to -3} \frac {(y + 3)(y - 4)}{(y + 3)}$

now what?

4. Originally Posted by Moo
For the first one, are you familiar with L'Hospital's rule ?
haha, we stand together! i am safe now. the mob won't hurt you, you can protect me

5. Thanks for the code.

And no, I am not familiar with that rule at all...

However, as you said to factor the one problem, i came up with an answer -7 by canceling two (y+3). According to my book, the answer is 0...

6. Originally Posted by Jhevon
haha, we stand together! i am safe now. the mob won't hurt you, you can protect me
You could've talked about the moob

Originally Posted by cjmac87
Thanks for the code.

And no, I am not familiar with that rule at all...

However, as you said to factor the one problem, i came up with an answer -7 by canceling two (y+3). According to my book, the answer is 0...
Hmmm well, I'm pretty sure the answer is -7... Did you look at the right answer ?

For the first one, do you know this formula :
$\displaystyle \lim_{h \to a} \frac{f(a+h)-f(a)}{h}=f '(a)$

Write your limit this way :

$\displaystyle \lim_{h \to 0} \frac{5^h-2^h}{h}$$\displaystyle =\lim_{h \to 0} \frac{(5^h\overbrace{-1)+(1}^{\text{this is 0}}-2^h)}{h}=\lim_{h \to 0} \frac{5^h-1}{h}-\lim_{h \to 0} \frac{2^h-1}{h}=\lim_{h \to 0} \frac{5^h-5^0}{h}-\lim_{h \to 0} \frac{2^h-2^0}{h} Apply the formula I gave you with a=0 and try to find the function f for the two limits 7. Originally Posted by cjmac87 Thanks for the code. And no, I am not familiar with that rule at all... L'Hopital's rule says if \displaystyle \lim_{x \to a} \frac {f(x)}{g(x)} \to \frac 00 \mbox{ or } \frac {\infty}{\infty}, then \displaystyle \lim_{x \to a} \frac {f(x)}{g(x)} = \lim_{x \to a} \frac {f'(x)}{g'(x)} However, as you said to factor the one problem, i came up with an answer -7 by canceling two (y+3). According to my book, the answer is 0... you are correct. the book is wrong, or you are looking in the wrong place 8. Originally Posted by Moo You could've talked about the moob oh yeah, if that's the case, we're both in trouble. just tell them you're the queen of MHF and so hurting you would cause an all out war. in addition, you have diplomatic immunity, so even the authorities have no business messing with you For the first one, do you know this formula : \displaystyle \lim_{h \to a} \frac{f(a+h)-f(a)}{h}=f '(a) Write your limit this way : \displaystyle \lim_{h \to 0} \frac{5^h-2^h}{h}$$\displaystyle =\lim_{h \to 0} \frac{(5^h\overbrace{-1)+(1}^{\text{this is 0}}-2^h)}{h}=\lim_{h \to 0} \frac{5^h-1}{h}-\lim_{h \to 0} \frac{2^h-1}{h}=\lim_{h \to 0} \frac{5^h-5^0}{h}-\lim_{h \to 0} \frac{2^h-2^0}{h}$

Apply the formula I gave you with a=0 and try to find the function f for the two limits
i thought of a similar way after i posted

9. Well, it's not that hard when actually having in mind a limit like $\displaystyle \underset{x\to 0}{\mathop{\lim }}\,\frac{e^{x}-1}{x}.$

Hence $\displaystyle \underset{h\to 0}{\mathop{\lim }}\,\frac{5^{h}-2^{h}}{h}=\underset{h\to 0}{\mathop{\lim }}\,\left( \frac{5^{h}-1}{h}-\frac{2^{h}-1}{h} \right).$

Now, consider $\displaystyle \underset{h\to 0}{\mathop{\lim }}\,\frac{5^{h}-1}{h}=\ln 5\cdot\underset{h\to 0}{\mathop{\lim }}\,\frac{e^{h\ln 5}-1}{h\ln 5}=\ln 5.$ In the same fashion we can find the value of the another limit, hence, both limits exist so we get that $\displaystyle \underset{h\to 0}{\mathop{\lim }}\,\frac{5^{h}-2^{h}}{h}=\underset{h\to 0}{\mathop{\lim }}\,\frac{5^{h}-1}{h}-\underset{h\to 0}{\mathop{\lim }}\,\frac{2^{h}-1}{h}=\ln \frac{5}{2}.$