heya guys!!!
solve it
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Lim x->0 ( a^x-a^-x/x)
If we put x in the expression we get value as 0/0 which is an indeterminate form.
Thus, we can use the L' Hospitals rule here.
Let's write the differentiated forms of both numerator and denominator separately.
Therefore,
lim x->0 a^x ln a - a^-x ( ln a) (-1) / 1
= Lim x->0 a^x ln a + a^-x ln a
= a^0 ( ln a) + a^-0 ( ln a)
= 2 ln a
Hope this helps you !
If we put x in the expression we get value as 0/0 which is an indeterminate form.
Thus, we can use the L' Hospitals rule here.
Let's write the differentiated forms of both numerator and denominator separately.
Therefore,
lim x->0 a^x ln a - a^-x ( ln a) (-1) / 1
= Lim x->0 a^x ln a + a^-x ln a
= a^0 ( ln a) + a^-0 ( ln a)
= 2 ln a
Hope this helps you !
Anjeelina:
2 log a
??
yes so you can write 2 ln a even as 2 log a
bcz the differentiation of a^x is not absolutely defined to be a^x ln a
I'm in, 11 standard tell me in mah language
it can even be a°x log a
ty
cleared now?
yss
alright
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Answer:
Lim x->0 ( a^x-a^-x/x)
If we put x in the expression we get value as 0/0 which is an indeterminate form.
Thus, we can use the L' Hospitals rule here.
Let's write the differentiated forms of both numerator and denominator separately.
Therefore,
lim x->0 a^x ln a - a^-x ( ln a) (-1) / 1
= Lim x->0 a^x ln a + a^-x ln a
= a^0 ( ln a) + a^-0 ( ln a)
= 2 ln a
Hope this helps you !
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