Verified Answer :
Answer:
0.25
Step-by-step explanation:
The given limit problem is
\(\lim_{h \to 0}\dfrac{\ln (4+h)-\ln (4)}{h}\)
On applying limit, we get
\(\dfrac{\ln (4+0)-\ln (4)}{0}=\dfrac{0}{0}\)
It is a 0/0 form.
Now, apply L' Hospital rule.
\(\lim_{h \to 0}\dfrac{\dfrac{d}{dh}[\ln (4+h)-\ln (4)]}{\dfrac{d}{dh}h}\)
\(\lim_{h \to 0}\dfrac{\dfrac{d}{dh}\ln (4+h)-\dfrac{d}{dh}\ln (4)}{\dfrac{d}{dh}h}\)
\(\lim_{h \to 0}\dfrac{\dfrac{1}{4+h}-0}{1}\)
\(\lim_{h \to 0}\dfrac{1}{4+h}\)
On applying limit, we get
\(\dfrac{1}{4+0}\)
\(\dfrac{1}{4}\)
\(0.25\)
Therefore, the value of given limit problem is 0.25.
The value of the expression is \(\dfrac{1}{4}\).
Given to us
- \(\lim_{h \to 0} \dfrac{(ln(4+h) - ln(4))}{h}\)
Limit as h approaches 0
As the limit h approaches 0,
\(\lim_{h \to 0} \dfrac{(ln(4+h) - ln(4))}{h}\)
substituting the values we get,
\(=\dfrac{(ln(4+0) - ln(4))}{0}\\\\=\dfrac{(ln(4) - ln(4))}{0}\\\\=\dfrac{0}{0}\)
As the expression is coming out to be in the form of \(\dfrac{0}{0}\), therefore, we will apply L'Hopital's rule.
L'Hopital's rule
\(\lim_{h \to 0} \dfrac{\dfrac{d}{dh}(ln(4+h) - \dfrac{d}{dh}ln(4))}{\dfrac{d}{dh}h}\)
\(=\lim_{h \to 0} \dfrac{\dfrac{1}{(4+h)} - 0}{1}\)
\(=\dfrac{\dfrac{1}{(4+0)}}{1}\)
\(=\dfrac{1}{4}\)
Hence, the value of the expression is \(\dfrac{1}{4}\).
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