Evaluate each of the following.
(a) lim x->0(e^x)-1-x/ x^2
(b) lim x->0 x-sinx/x^3
(c) lim x->infinity (In x)^2/x
(d) lim x->0+ (sinx)In x
(e) lim x->0+ (cos3x)^5/x
(f) lim x->1+ ((1/x-1) -(1/In x))
b.lim 1-1/3cosx^2
To evaluate each of the given limits, we will go step by step and apply different approaches to find the limit.
(a) lim x->0 (e^x) - 1 - (x/x^2):
To evaluate this limit, we can use L'Hopital's rule. We can differentiate the numerator and denominator with respect to x and then calculate the limit again.
Differentiating the numerator:
d/dx (e^x - 1 - x) = e^x - 1.
Differentiating the denominator:
d/dx (x^2) = 2x.
Calculating the limit of the derivative:
lim x->0 (e^x - 1)/(2x).
Now, we can apply L'Hopital's rule again, differentiating the numerator and denominator again and calculating the limit:
Differentiating the numerator:
d/dx (e^x - 1) = e^x.
Differentiating the denominator:
d/dx (2x) = 2.
Calculating the limit of the derivative:
lim x->0 e^x/2 = e^0/2 = 1/2.
Therefore, the limit of lim x->0 (e^x) - 1 - (x/x^2) is 1/2.
(b) lim x->0 (x - sin(x))/x^3:
To evaluate this limit, we can use Taylor series expansion. The Taylor series expansion for sin(x) is sin(x) = x - x^3/3! + x^5/5! -...
Substituting this expansion into the limit expression:
lim x->0 (x - (x - x^3/3! + x^5/5! - ...))/x^3.
Simplifying the expression:
lim x->0 (x - x + x^3/3! - x^5/5! + ... )/x^3.
We can see that all terms except x^3/3! are canceled as x approaches 0. Therefore, the limit becomes:
lim x->0 x^3/3! / x^3 = 1/3!.
Therefore, the limit of lim x->0 (x - sin(x))/x^3 is 1/3!.
(c) lim x->infinity ((ln(x))^2/x):
To evaluate this limit, we can use the property of logarithmic functions. Logarithmic functions grow slower than polynomial functions as x approaches infinity.
Therefore, the limit becomes:
lim x->infinity (log(x)^2)/x = 0.
Therefore, the limit of lim x->infinity ((ln(x))^2/x) is 0.
(d) lim x->0+ (sin(x))(ln(x)):
To evaluate this limit, we can apply L'Hopital's rule. Differentiating both the numerator and the denominator with respect to x:
Differentiating the numerator:
d/dx (sin(x)) = cos(x).
Differentiating the denominator:
d/dx (ln(x)) = 1/x.
Calculating the limit of the derivative:
lim x->0+ cos(x)/(1/x).
Since cos(x) is bounded between -1 and 1, the limit can be rewritten as:
lim x->0+ (1/x) = +infinity.
Therefore, the limit of lim x->0+ (sin(x))(ln(x)) is +infinity.
(e) lim x->0+ (cos(3x))^5/x:
To evaluate this limit, we can use the property of trigonometric functions. The cosine function oscillates between -1 and 1 as x approaches 0.
Therefore, the limit becomes:
lim x->0+ (1^5)/x = +infinity.
Therefore, the limit of lim x->0+ (cos(3x))^5/x is +infinity.
(f) lim x->1+ ((1/(x-1)) - (1/ln(x))):
To evaluate this limit, we can substitute the value of x into the expression and see what it approaches from the right side:
lim x->1+ ((1/(x-1)) - (1/ln(x))).
Let's substitute x=1 into the expression:
((1/(1-1)) - (1/(ln(1)))) = (1/0 - 1/(-infinity)).
Here, the denominator of the first term is 0, and the denominator of the second term approaches negative infinity.
Therefore, the limit of lim x->1+ ((1/(x-1)) - (1/ln(x))) is undefined.