Advertisements
Advertisements
प्रश्न
Evaluate the following Limits: `lim_(x -> 0)[("a"^x + "b"^x + "c"^x - 3)/x]`
Advertisements
उत्तर
`lim_(x -> 0)[("a"^x + "b"^x + "c"^x - 3)/x]`
= `lim_(x -> 0) (("a"^x - 1) + ("b"^x - 1) + ("c"^x - 1))/x`
= `lim_(x -> 0) (("a"^x - 1)/x + ("b"^x - 1)/x + ("c"^x - 1)/x)`
= `lim_(x -> 0)(("a"^x - 1)/x) + lim_(x -> 0)(("b"^x - 1)/x) + lim_(x -> 0)(("c"^x - 1)/x)`
= log a + log b + log c ...`[lim_(x -> 0) ("a"^x - 1)/x = log "a"]`
= log (abc).
APPEARS IN
संबंधित प्रश्न
Evaluate the following: `lim_(x -> 0) [(2^x - 1)^2/((3^x - 1) xx log (1 + x))]`
Evaluate the following Limits: `lim_(x -> 0)[(5^x - 1)/x]`
Evaluate the following Limits: `lim_(x -> 0)[((5^x - 1)^2)/(x*log(1 + x))]`
Evaluate the following limit :
`lim_(x -> 0) [(6^x + 5^x + 4^x - 3^(x + 1))/sinx]`
Evaluate the following limit :
`lim_(x -> 0) [(log(3 - x) - log(3 + x))/x]`
Evaluate the following limit :
`lim_(x -> 0) [(5 + 7x)/(5 - 3x)]^(1/(3x))`
Select the correct answer from the given alternatives.
`lim_(x -> 0) [(log(5 + x) - log(5 - x))/sinx]` =
Select the correct answer from the given alternatives.
`lim_(x -> pi/2) ((3^(cosx) - 1)/(pi/2 - x))` =
Evaluate the following :
`lim_(x -> 1) [("ab"^x - "a"^x"b")/(x^2 - 1)]`
Evaluate the following :
`lim_(x -> 0) [((5^x - 1)^2)/((2^x - 1)log(1 + x))]`
`lim_{x→∞} ((3x + 3)^40(9x - 3)^5)/(3x + 1)^45` = ______
lf the function f(x) satisfies `lim_{x→1}(2f(x) - 5)/(2(x^2 - 1)) = e`, then `lim_{x→1}f(x)` is ______
`lim_(x -> 0) (sin^4 3x)/x^4` = ________.
Evaluate the following:
`lim_(x->0)[((25)^x - 2(5)^x + 1)/x^2]`
Evaluate the following limit :
`lim(x>2)[(z^2 -5z+6)/(z^2-4)]`
Evaluate the following limit :
`lim_(x->0)[(sqrt(6+x+x^2)-sqrt6)/x]`
Evaluate the following:
`lim_(x -> 0)[((25)^x - 2(5)^x + 1)/x^2]`
Evaluate the following:
`lim_(x->0)[((25)^x - 2(5)^x + 1)/x^2]`
