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Question

Question: Evaluate the following: \(\displaystyle \lim_{x \to 0}\left[ \dfrac{\log 100+\log \left( 0.01+x \rig...

Evaluate the following: limx0[log100+log(0.01+x)x]\displaystyle \lim_{x \to 0}\left[ \dfrac{\log 100+\log \left( 0.01+x \right)}{x} \right]?

Explanation

Solution

The above given function is limx0[log100+log(0.01+x)x]\displaystyle \lim_{x \to 0}\left[ \dfrac{\log 100+\log \left( 0.01+x \right)}{x} \right]. Here we have to find out the limit of the given function. a limit is defined as a number approached by the function as an independent function`s variable approaches a particular value. For example, the functionf(x)=4xf\left( x \right)=4x, the limit of f(x)f\left( x \right) as x approaches 22, then f(2)=8f\left( 2 \right)=8.

Complete step by step solution:
The given function is:
f(x)=limx0[log100+log(0.01+x)x]\Rightarrow f\left( x \right)=\displaystyle \lim_{x \to 0}\left[ \dfrac{\log 100+\log \left( 0.01+x \right)}{x} \right]
We have to find the limit of the given function when x approaches to 0.
Now, first we will simplify the above given function. For this we will use logarithm formula, which is log(m,n)=logm+logn\log \left( m,n \right)=\log m+\log n, so we will apply this formula on the above log100+log(0.01+x)\log 100+\log \left( 0.01+x \right), then we get
f(x)=limx0[log100+log(0.01+x)x] f(x)=limx0[log100(0.01+x)x] \begin{aligned} & \Rightarrow f\left( x \right)=\displaystyle \lim_{x \to 0}\left[ \dfrac{\log 100+\log \left( 0.01+x \right)}{x} \right] \\\ & \Rightarrow f\left( x \right)=\displaystyle \lim_{x \to 0}\left[ \dfrac{\log 100\left( 0.01+x \right)}{x} \right] \\\ \end{aligned}
Now multiply 100100 with (0.01+x)\left( 0.01+x \right) , then we get
f(x)=limx0[log(1+100x)x]\Rightarrow f\left( x \right)=\displaystyle \lim_{x \to 0}\left[ \dfrac{\log \left( 1+100x \right)}{x} \right]
Now multiply and divide the above expression by 100100, then we get
f(x)=limx0[log(1+100x)100x]×100\Rightarrow f\left( x \right)=\displaystyle \lim_{x \to 0}\left[ \dfrac{\log \left( 1+100x \right)}{100x} \right]\times 100
Now if we apply limit on x then, as x approaches to zero the term100x100x also approaches to zero, and we also know that
f(x)=limx0[log(1+x)x]=1\Rightarrow f\left( x \right)=\displaystyle \lim_{x \to 0}\left[ \dfrac{\log \left( 1+x \right)}{x} \right]=1 by simply writing the expansion oflog(1+x)\log \left( 1+x \right).
Thus if100x100xapproaches to zero the value off(x)=limx0[log(1+100x)100x]f\left( x \right)=\displaystyle \lim_{x \to 0}\left[ \dfrac{\log \left( 1+100x \right)}{100x} \right] also equal to one, the putting this in the expressionf(x)=limx0[log(1+100x)100x]×100f\left( x \right)=\displaystyle \lim_{x \to 0}\left[ \dfrac{\log \left( 1+100x \right)}{100x} \right]\times 100 , then we get
f(x)=limx0[log(1+100x)100x]×100 f(x)=1×100 f(x)=100 \begin{aligned} & \Rightarrow f\left( x \right)=\displaystyle \lim_{x \to 0}\left[ \dfrac{\log \left( 1+100x \right)}{100x} \right]\times 100 \\\ & \Rightarrow f\left( x \right)=1\times 100 \\\ & \Rightarrow f\left( x \right)=100 \\\ \end{aligned}
Hence the limit of the given expression is 100100.

Note: To solve these types of questions we have to simplify them, if we directly put the limit then it becomes hard to get the correct value. Here to solve the above expression we used the logarithm formula log(m,n)=logm+logn\log \left( m,n \right)=\log m+\log n, sometimes the question may be different then we have to use any other formula by which we can simplify it.