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**1. Suppose that f is differentiable on R and [tex]\lim_{x \rightarrow \infty} f'(x) [/tex] = M. Show that [tex]\lim_{x \rightarrow \infty} (f(x+1)-f(x)) [/tex] also exists, and compute it.**

**3. I am pretty sure the limit will be equal to m. Here is my attempt.**

[tex]\lim_{x \rightarrow \infty} f'(x) [/tex] = [tex]\lim_{x \rightarrow \infty} \lim_{h

\rightarrow 0} \frac{f(x+h)-f(x)}{h} [/tex]

By continuity this is equal to:

[tex]\lim_{h \rightarrow 0} \lim_{x \rightarrow \infty} \frac{f(x+h)-f(x)}{h} [/tex]

Fixing h=1

This is equal to [tex]\lim_{h \rightarrow 0} \lim_{x \rightarrow \infty} \frac{f(x+1)-f(x)}{1} [/tex]

=[tex]\lim_{x \rightarrow \infty} (f(x+1)-f(x)) [/tex].

Have I done this right? I don't think you can just swap the limits like that can you? Does anybody have any suggestions?[tex]\lim_{x \rightarrow \infty} f'(x) [/tex] = [tex]\lim_{x \rightarrow \infty} \lim_{h

\rightarrow 0} \frac{f(x+h)-f(x)}{h} [/tex]

By continuity this is equal to:

[tex]\lim_{h \rightarrow 0} \lim_{x \rightarrow \infty} \frac{f(x+h)-f(x)}{h} [/tex]

Fixing h=1

This is equal to [tex]\lim_{h \rightarrow 0} \lim_{x \rightarrow \infty} \frac{f(x+1)-f(x)}{1} [/tex]

=[tex]\lim_{x \rightarrow \infty} (f(x+1)-f(x)) [/tex].

Have I done this right? I don't think you can just swap the limits like that can you? Does anybody have any suggestions?

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