Question

Let \(f(x)\) be differentiable at \(x=h\) then $\lim _{x \rightarrow h} \frac{(x+h) f(x)-2 h f(h)}{x-h}$ is equal to (A) \(f(h)+2 h f^{\prime}\) (h) (B) \(2 f(h)+h f^{\prime}\) (C) \(\mathrm{hf}(\mathrm{h})+2 \mathrm{f}^{\prime}\) (h) (D) \(\mathrm{hf}(\mathrm{h})-2 \mathrm{f}^{\prime}\) (h)

Short answer

Answer: (A) \(f(h) + 2hf'(h)\)

Step-by-step solution

Dealing with limit

In this step, we will break the given expression into parts and use limit properties to simplify them. Let's rewrite the given limit expression as follows: $$ \lim_{x\rightarrow h} \frac{(x+h) f(x)-2hf(h)}{x-h} $$ Now, let's separate the terms in the numerator: $$ \lim_{x\rightarrow h} \frac{(xf(x)+hf(x))-2hf(h)}{x-h} $$ We can now split the limit into two different limits: $$ \lim_{x\rightarrow h} \frac{xf(x)}{x-h} + \lim_{x\rightarrow h} \frac{hf(x)-2hf(h)}{x-h} $$

Finding the limits separately

Now, we need to find the limits of the two expressions: 1. For the first limit, we need to find $$ \lim_{x\rightarrow h} \frac{xf(x)}{x-h} $$ Before we proceed, let's use the property of limits that states if the limit of a product exists, it is equal to the product of their limits. Thus, $$ \lim_{x\rightarrow h} xf(x) = x\lim_{x\rightarrow h} f(x) $$ So, as x approaches h, the limit becomes: $$ h\lim_{x\rightarrow h} f(x) = hf(h) $$ Now, $$ \lim_{x\rightarrow h} \frac{xf(x)}{x-h} = \lim_{x\rightarrow h} \frac{hf(h)}{x-h} = f(h) $$ 2. For the second limit, we need to find $$ \lim_{x\rightarrow h} \frac{hf(x)-2hf(h)}{x-h} $$ Factor the constant h from the numerator and apply the limit property: $$ h\lim_{x\rightarrow h} \frac{f(x)-2f(h)}{x-h} $$ Now, let's use the definition of the derivative of \(f(x)\) at \(x=h\): $$ f'(h) = \lim_{x\rightarrow h} \frac{f(x)-f(h)}{x-h} $$ Then we have: $$ h\lim_{x\rightarrow h} \frac{f(x)-2f(h)}{x-h} = h\left(2f'(h)\right) = 2hf'(h) $$

Combining the limits and finding the final answer

In this step, we will combine the limits we found in Step 2. Thus, $$ \lim_{x\rightarrow h} \frac{(x+h) f(x)-2 hf(h)}{x-h} = f(h) + 2hf'(h) $$ Now, we can compare our result with the given choices and see that it matches choice (A). Therefore, the correct answer is (A) \(f(h) + 2hf'(h)\).