lim x->inf (1+x^2)e^-x = lim x->inf (1+x^2) / e^x Differentiate denominator and numerator (L Hospital Rule) lim x->inf (2x / e^x) Again apply L Hospital Rule, lim x->inf (2 / e^x) put x=infinity and the expression would become 0. (since 1/infinity = 0) So, the correct option is (A) 0.
Lim x->infinity (1+x^2)e^-x
ReplyDeletea)0
b)1
c)1/2
d)infinity
lim x->inf (1+x^2)e^-x = lim x->inf (1+x^2) / e^x
ReplyDeleteDifferentiate denominator and numerator (L Hospital Rule)
lim x->inf (2x / e^x)
Again apply L Hospital Rule,
lim x->inf (2 / e^x)
put x=infinity and the expression would become 0. (since 1/infinity = 0)
So, the correct option is (A) 0.
Thanks Akshay for posting the question.
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ReplyDeleteits lim x->inf (1+x^2)^e^x
ReplyDeleteSo ans : 1
Yes, you are correct.
DeleteIf question is :
lim x->inf(1+x^2)^e^x
Then answer would be 1.
Detailed Solution-
Let us assume,
f(x) = (1+x^2) and g(x) = e^x
So it's a form of,
lim x->inf f(x)^g(x)
Now, we know, a^b = exp (b ln a)
so, lim x->inf f(x)^g(x) = lim x->inf exp(g(x).ln f(x))
=> exp (lim x->inf g(x).ln f(x) )
=> exp (lim x->inf e^x.ln(1+x^2))
After applying L hospital rule and putting the limit inside the brackets, the expression inside the bracket becomes 0.
Now, exp(0) = 1
Let me know in case of any query.