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Question: Find \(\mathop {\lim }\limits_{x \to 0} \;\dfrac{{\ell n(2 + x) + \ell n\;0.5}}{x}\)...

Find lim⁡x→0  ℓn(2+x)+ℓn  0.5x\mathop {\lim }\limits_{x \to 0} \;\dfrac{{\ell n(2 + x) + \ell n\;0.5}}{x}

Explanation

Solution

In this question we are asked to solve a function with limit. In order to proceed with this question we need to know the L'Hospital Rule. L'Hospital's rule is the definitive way to simplify functions which has limits. It does not directly evaluate limits, but simplifies evaluation. An indeterminate form is an expression involving two functions whose limit cannot be determined only from the limits of the individual functions. Intermediate means unknown value. The value of function cannot be calculated even after putting the limits.

Complete step by step answer:
We are given,
lim⁡x→0  ℓn(2+x)+ℓn  0.5x\mathop {\lim }\limits_{x \to 0} \;\dfrac{{\ell n(2 + x) + \ell n\;0.5}}{x}
If we put the value asx=0x = 0 then the fraction should be in intermediate form i.e. should become 00  \dfrac{0}{0}\; nor become∞∞\dfrac{\infty }{\infty } , but the numerator is not tending to 0 or ∞\infty ,so we’ll have to further simplify it.
⇒lim⁡x→0  ℓn(2+x)+ℓn  (12)x\Rightarrow \mathop {\lim }\limits_{x \to 0} \;\dfrac{{\ell n(2 + x) + \ell n\;(\dfrac{1}{2})}}{x}

⇒lim⁡x→0  ℓn(2+x)+ℓn  1−ℓn  2x \Rightarrow \mathop {\lim }\limits_{x \to 0} \;\dfrac{{\ell n(2 + x) + \ell n\;1 - \ell n\;2}}{x}
Now this function is in indeterminate form, and becomes in the form 00\dfrac{0}{0}
We’ll apply the L'Hospital Rule, and differentiate the function.
⇒lim⁡x→0  12+x×1+0−01\Rightarrow \mathop {\lim }\limits_{x \to 0} \;\dfrac{{\dfrac{1}{{2 + x}} \times 1 + 0 - 0}}{1}
⇒lim⁡x→0  12+x\Rightarrow \mathop {\lim }\limits_{x \to 0} \;\dfrac{1}{{2 + x}}
Now if we putx=0x = 0
We’ll get,
⇒12\Rightarrow \dfrac{1}{2}
This is the required answer.

Note: L’Hospital Rule can only in applied if direct substitution leads to indeterminate form. If the problem is out of the indeterminate forms, you can’t be able to apply the L'Hospital Rule. Before applying L'Hospital Rule make sure both, the numerator and denominator either tend to 0 or ∞\infty . If one of the both does not tend to be 0 or ∞\infty , then L'Hospital Rule can’t be applied.
L’Hospital Rule can only be applied if –
If the function can be differentiated
If the function has a limit