# Thread:

1. ## I have a problem in understanding a problem about the limit of a function.

The question is as follows:

Use l'Hopital's rule to evaluate

lim [(a^x - b^x) / (e^ax - e^bx)] when x approaches 0. (Notice: e here stands for exponential symbol).

The answer given by the textbook is (lna - lnb) / (a - b). However, I have problem of getting to this answer. In particular, I don't know how to obtain (a - b) in the denominator.

Please show me the steps of getting to answer. thank you in advance

2. Originally Posted by Real9999
The question is as follows:

Use l'Hopital's rule to evaluate

lim [(a^x - b^x) / (e^ax - e^bx)] when x approaches 0. (Notice: e here stands for exponential symbol).

The answer given by the textbook is (lna - lnb) / (a - b). However, I have problem of getting to this answer. In particular, I don't know how to obtain (a - b) in the denominator.

Please show me the steps of getting to answer. thank you in advance

to where did u come with your work, show and we'll see where is the problem that u have

just to note that :
$\displaystyle \frac {d}{dx} ( e^{ax} )=ae^{ax}$

$\displaystyle \frac {d}{dx} ( a^{x} )=a^{x} \ln {a}$

$e^0 =1$

does this helps ? or ? need to go in that more detailed ?

anyway after using L'Hospital's rule (and knowing derivate's from above ) u can easy come to conclusion that result will be

$\displaystyle \frac {\ln{a}-\ln{b}}{a-b}$

3. this can be done by not using that rule, if you want the solution, let me know.

4. Originally Posted by Krizalid
this can be done by not using that rule, if you want the solution, let me know.
but it's clearly stated that he have to use L'Hospital's rule by his concept of question : "Use L'Hospital's rule to evaluate.. " and evaluating this using any another procedure could (although lead to the correct solution) lead to incorrect solution (as for teachers point of view)

5. Originally Posted by yeKciM
but it's clearly stated that he have to use L'Hospital's rule by his concept of question : "Use L'Hospital's rule to evaluate.. " and evaluating this using any another procedure could (although lead to the correct solution) lead to incorrect solution (as for teachers point of view)
Which is why he did not provide the solution but rather asked him if he wants the other solution :P

6. Thank you very much for your replies! thanks to all! And sorry for the late reply as I was sick over the past two days...

I still have a problem in understanding how to come to the (a-b) in the denominator. My procedure of doing this question is to differentiate the original function, [(a^x - b^x) / (e^ax - e^bx)], by using the quotient rules. I came to (e^ax - e^bx)^2, and then to (e^ax - e^bx) in the denominator after the entire calculation by the quotient rule. That is, I still do not understand how to go from (e^ax - e^bx) to (a - b).

Sorry for the mess, I will now learn how to type mathematical symbols.

7. ## Still have a problem.

Oh, I know how to come to the answer now! The problem I had was that I misunderstood the l'Hopital's rule!!! thanks to all!
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 4, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9653019309043884, "perplexity_flag": "middle"}