Question

$ lim_{ h \to 0 } \frac{ f (2 h + 2 + h^2 ) - f \, (2)}{ f \, ( h - h^2 + 1) - f (1)} $, given thta f ' (2) = 6 and f ' (1) = 4.

• A. does not exist
• B. is equal to -3/2
• C. is equal to 3/2
• D. is equal to 3

Answer 1

The Correct Option is A

Here, $ lim_{ h \to 0 } \frac{ f (2 h + 2 + h^2 ) - f \, (2)}{ f \, ( h - h^2 + 1) - f (1)} $
$\hspace 25mm$ [ $ \because $ f ' (2) = 6 and f ' (1) = 4, given ]
Applying L'Hospital's rule,
= $ lim_{ h \to 0 } \frac{ \{ f \, ' (2 h + 2 + h^2 ) \} . (2 + 2h) - 0 }{ \{ f \, ' ( h - h^2 + 1 ) \} . (1 - 2h) - 0 } = \frac{ f \, ' (2) . 2 }{ f \, ' \, (1) . 1 } $
= $ \frac{ 6 . 2}{ 4. 1} = 3 $ $\hspace15mm$ [using f ' (2) = 6 and f ' (1) = 4]

Answer 2

Ans. For assessing indeterminate forms like 0/0 or ∞/∞, use the L'Hospital rule. L'Hospital's theorem is used in calculus to assess the limits of derivatives with indeterminate forms. This rule can be used several times. This rule keeps an indefinite form after each application, even if we only use it once. The hospital rule won't assist, though, if the issue isn't with one of the uncertain types.

If we want to make use of this regulation, we must first ensure that the limit is in the correct format. This is accomplished in the following manner:

To use this rule, we must ensure that the fraction is made up of two functions, f(x)/g (x)

It is critical to note that when the x-value is entered, the function must evaluate to either 0/0 or ∞/∞, as these are the two sorts of indeterminate forms. If the limit problem is not indeterminate, we won't be able to utilize this method directly.

The L’Hospital Rule is given by the formula,

When direct substitution of a limit generates an indeterminate form, we can use L'Hospital's rule.

The rule of L'Hospital is as follows:

The limit of a quotient of functions (i.e., an algebraic fraction) equals the limit of their derivatives.

It's vital to notice that L'Hopital's rule does not apply the quotient rule and treats f(x) and g(x) as independent functions.