# Thread:

1. ## Pre-Calculus Question w/ Some Basic Calculus Involved

Hi,

$\lim_{\substack{x-3}} f(x)=\frac{x^2+4x-21}{x-3}$.

Using L'Hopital's rule (or algebra manipulation) we can determine the limit is 10. My question is, how is this possible if it is a rational function with a vertical asymptote at 3? Shouldn't it approach infinity?

After graphing it, I noticed that it happens to be a straight line and virtually equal to the graph of its derivative $f'(x)=2x+4$? This seems odd, as the graph said there was a hole at the point x=3, not a vertical asymptote. If I slightly change the function (such as x-4 in the denominator), it looks more like a rational function with a vertical asymptote.

I don't need any computational help with this problem - I just want to understand what's going on here.

2. Nevermind - I realized after factoring the numerator that the factor (x-3) from the denominator cancels out, essentially creating the function y=x+7 with a hole at x=3, equivelent to the entire function when you graph it.

This is what I get for teaching myself calculus while only just starting pre-calculus.
{"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": 2, "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": 15, "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": 0.9344742894172668, "perplexity_flag": "head"}