limx→∞lnx+√x2+1x+√x2−1⋅(lnx+1x−1)−2=18
Any suggestion to find this limit without series expansion and l'Hôpital's rule? Thanks and regards.
Note:
WolframAlpha confirms that the result is 18.
Answer
It is easy to observe that as x→∞ the argument of both the logarithms tend to 1 and this is exactly what we need when we are dealing with limit of logarithmic terms. Just add 1 and subtract 1 in the argument of log and proceed as follows:
L=limx→∞logx+√x2+1x+√x2−1(logx+1x−1)−2=limt→0+log1+√1+t21+√1−t2(log1+t1−t)−2 (putting x=1/t)=limt→0+log(1+1+√1+t21+√1−t2−1){log(1+1+t1−t−1)}−2=limt→0+log(1+√1+t2−√1−t21+√1−t2){log(1+2t1−t)}−2=limt→0+log(1+√1+t2−√1−t21+√1−t2)√1+t2−√1−t21+√1−t2⋅√1+t2−√1−t21+√1−t2{log(1+2t1−t)}−2=limt→0+√1+t2−√1−t21+√1−t2{log(1+2t1−t)2t1−t⋅2t1−t}−2=limt→0+√1+t2−√1−t21+√1−t2⋅(1−t)24t2=18limt→0+√1+t2−√1−t2t2=18limt→0+2t2t2{√1+t2+√1−t2}=18
The technique used above is based on the standard limits and is well explained in the comments to this answer.
No comments:
Post a Comment