21OH!!!!!1
Yah!! I had taken -∞ * ∞ as indeterminate form and workd out...........
Thnx Sir [1]
1)\lim_{x->0}xlogsinx \; 2) \lim_{x->0} (cosecx)^{^{1/logx}} 3)\lim_{x->0} ((1+x)^{^{1/x}}-e)/x 4)\lim_{x->0}[(sinx)^{1/x}+ (1/x)^{sin x}]
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62log sin x
1/x
= cos x /sin x
-1/x2
= -x2 cos x /sin x
yes it should be zero! not infinity!
21Oh yes!!!
Sir CAn v expect questns in JEE wer v USE expansion Series of log/sin/cos/e...... etc. That is in PHYSICS N MATHS both
663)
\lim_{x\to 0}\dfrac{(1+x)^{1/x}-e}{x}=\lim_{x\to 0} (1+x)^{1/x}\ \dfrac{x-\ln(1+x)-x\ln(1+x)}{x^2(1+x)}
The limit of (1+x)1/x as x→0 is quite obviously e. So it remains to evaluate \lim_{x\to 0} \dfrac{x-\ln(1+x)-x\ln(1+x)}{x^2(1+x)}
Use the series expansion of ln(1+x), namely
\ln(1+x)=x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\ldots
We get the numerator (after some simplification) as
-\dfrac{x^2}{2}+ \textrm{terms containing }x^3 \textrm{ and higher powers}
Accordigly when we divide by the denominator, the expression becomes
\lim_{x\to 0} \dfrac{-\dfrac{1}{2}+ \textrm{terms containing }x \textrm{ and higher powers} }{1+x}= -\dfrac{1}{2}
Accordingly, the required limit becomes -\dfrac{e}{2}
21664) You can individually apply the limits to both summands because the limit exists for both.
Let A = (\sin x)^{1/x}
Taking natural log on both side and using the continuity of log, we get
\ln(\lim_{x\to 0} A) = \dfrac{1}{x}\ \ln\sin x
As x->0, the right side is turning out to be ∞ X -∞ form, which is not an indeterminate. Accordingly, we get
\ln(\lim_{x\to 0} A) = -\infty
implying that limx→0 A = 0
The next term,
B=\left(\dfrac{1}{x}\right)^\sin x
It is ∞0 form which is indeterminate, accordingly we get
\ln \left(\lim_{x\to0} B\right) =-\lim_{x\to0}\ \sin x\ln x = - \lim_{x\to 0}\dfrac{\ln x}{\csc x} = -\lim_{x\to 0} \dfrac{1/x}{-\csc x\cot x}=0
Hence, limx→0B=1
21Mani i had dun it that way only...... i got 1
But see #17
Apply LH in ther....... i m gting 2 by that methd.......
mayb sum calc err
21nishant sir, i applied LH rule in first ques but the ans coming is 0 while ans given is infinity?????
62For Q2: \lim_{x\rightarrow 0}\frac{log cosec x}{\log x} \\=\lim_{x\rightarrow 0}\frac{\frac{d}{dx}log(cosec x)}{\frac{d}{dx}\log x} \\=\lim_{x\rightarrow 0}-\frac{cosec x \times cot x/ cosec x}{1/x} \\=\lim_{x\rightarrow 0}-\frac{x}{tan x}
62LH is L Hospital rule!
It may be out of syllabus but it is a must know
You cannot not know it!
1L'Hospital rule... :)
waise, it's out of jee syllabus... lekin objective pattern mein kaun dekhtha hain... [3]