Can I grep only the first n lines of a file?

The magic of pipes;

head -10 log.txt | grep <whatever>

Or use awk for a single process without |:

awk '/your_regexp/ && NR < 11' INPUTFILE

On each line, if your_regexp matches, and the number of records (lines) is less than 11, it executes the default action (which is printing the input line).

Or use sed:

sed -n '/your_regexp/p;10q' INPUTFILE 

Checks your regexp and prints the line (-n means don't print the input, which is otherwise the default), and quits right after the 10th line.

You have a few options using programs along with grep. The simplest in my opinion is to use head:

head -n10 filename | grep ...

head will output the first 10 lines (using the -n option), and then you can pipe that output to grep.

For folks who find this on Google, I needed to search the first n lines of multiple files, but to only print the matching filenames. I used

 gawk 'FNR>10 {nextfile} /pattern/ { print FILENAME ; nextfile }' filenames

The FNR..nextfile stops processing a file once 10 lines have been seen. The //..{} prints the filename and moves on whenever the first match in a given file shows up. To quote the filenames for the benefit of other programs, use

 gawk 'FNR>10 {nextfile} /pattern/ { print "\"" FILENAME "\"" ; nextfile }' filenames